Mixing
Help save this proof about the regularity of the lebesgue measure on $mathbb{R}^d$
$begingroup$
In a previous homework our task was to prove the regularity of the Lebegue-measure on $mathbb{R}^d$.
More precisely:
Let $(mathbb{R}^d, mathcal{M}^*, lambda)$ be a measure space and $mathcal{M}^* := mathcal{M}^*(mathbb{R}^d)$ be the $sigma$-algebra of the $lambda^*$-measurable sets, with $lambda$ being the by the Lebesgue-outer-measure $lambda^*$ induced measure.
Show that
begin{align*}
lambda(A)
& = inf{lambda(O) mid O subset mathbb{R}^d text{ is open and } A subset O } \
& = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset A }.
end{align*}
Our attempt goes as follows:
For all $A in mathcal{M}^*$ we want to show
begin{equation*}
lambda(A)
ge inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
ge sup{ lambda(K): K subset mathbb{R}^d text{ compact and } K subset A }
ge lambda(A)
end{equation*}
First inequality:
If $lambda(A) = infty$, because of the monotonicity of the measure, we have $lambda(O) = infty$ for all open sets $O$ with $A subset O$.
If $lambda(A) < infty$, we define $A_{delta} := bigcup_{a in A} U_{delta}(a)$, which is a open and therefore measurable set with $A subset A_{delta}$.
Now we have $A_{delta} setminus A xrightarrow{delta to 0} emptyset$ and because of the continuity from below in the emptyset of $lambda$, and, because $A_{delta} setminus A$ is measurable, because it's the difference of measurable sets,
begin{equation*}
forall varepsilon > 0
exists delta > 0:
lambda(A_{delta} setminus A) < varepsilon.
end{equation*}
Because $A_{delta}$ is open and $sigma$-additivity of $lambda$ we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(A_{delta})
= lambda(A_{delta} setminus A) + lambda(A)
< varepsilon + lambda(A).
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
The second inequality follows from the monotonicity of $lambda^*$:
From $K subset A subset O$ we have $lambda(K) le lambda(O)$ for $K$ und $O$ as defined above.
Because taking the supremum or infimum doesn't change weak inequalities, the inequality follows.
The third inequality:
One can show, that every open set is $sigma$-compact.
Lemma
Let $A_{delta} subset mathbb{R}^d$ be an open subset.
Then there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$, so that $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Proof
Let $a in A_{delta} $.
Because $A_{delta} $ is open, there exists an $varepsilon_{a} > 0$, so that $U_{varepsilon_a}(a) subset A_{delta} $.
For every $a in A_{delta} $ we choose a bounded closed axially parallel cube $W_a$ with rational midpoint from $mathbb{Q}^d$ and rational edge length $q in mathbb{Q}$, so that
begin{equation*}
a
in W_{a}
subset U_{varepsilon_a}(a)
subset A_{delta} .
end{equation*}
This is always possible, because $mathbb{Q}$ is dense in $mathbb{R}$.
Therefore, we have $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Since there are only countable many cubes of this form, the union is countable. $square$
We let $K_n := bigcup_{j = 1}^{n} W_j$, which is compact as union of compact sets.
Then $K_n xrightarrow{n to infty} A_{delta}$.
Now we have $A_{delta} setminus K_n xrightarrow{n to infty} emptyset$ and with analogous argumentation as above
begin{equation*}
forall varepsilon > 0
exists N in mathbb{N}:
lambda(A_{delta} setminus K_n) < varepsilon
forall n > N.
end{equation*}
Therefore follows for all $varepsilon > 0$
begin{align*}
lambda(A)
le lambda(A_{delta})
le lambda(A_{delta}setminus K_n) + lambda(K_n)
< varepsilon + sup{ lambda(K): K subset mathbb{R}^d text{ compact und } K subset A_{delta} },
end{align*}
Because $varepsilon > 0$ was arbitrary and $delta$ can be arbitrarily small, the inequality follows.
My problem
I assume the proof of the second inequality is right. But: I not sure if the reasoning at the end of the proof of the last inequality is rigorous enough. Especially, if you take $A := mathbb{Q}$ in the proof of the first inequality, then for every $delta > 0$, we have $A_{delta} = mathbb{R}$ and therefore, we don't have $A_{delta} setminus A to emptyset$ or even $lambda(A_{delta} setminus A) to 0$.
Is there anyway to ''save'' this proof by fixing it and not changing the approach?
Correct Proof
First inequality
Case 1: $lambda(A) = infty$.
As above.
Case 2: $lambda(A) < infty$.
Utilising the Caratheodory construction of the Lebesgue measure, we know that
$$
forall varepsilon > 0
exists (a_n,b_n] := prod_{i=1}^d left(a_{n}^{(i)},b_{n}^{(i)}right]:
A subset bigcup_{n in mathbb{N}} (a_n,b_n]
quad text{and} quad
sum_{k=1}^infty lambda left((a_k,b_k]right)
< lambda(A) + frac{varepsilon}{2},
$$
where the $d$-dimensional cubes $(a_n,b_n]$ are pairwise disjoint.
Now let
$$
U := bigcup_{n=1}^infty (a_n, b_n+ t_n varepsilon)
qquad text{with} qquad t_n := 2^{-n-2d-1} max{1,b_{n}^{(i)} -a_{n}^{(i)}}^{-(d-1)}.
$$
Now we have $A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and $lambda(U) < lambda(A) + varepsilon$.
Therefore, we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(U)
< lambda(A) + varepsilon.
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
Second inequality
Same as above
Third inequality
From a previous homework we know that $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$ and therefore, that
begin{equation*}
forall M in mathcal{M}^*
exists B in mathcal{B}(mathbb{R}^d), N in mathcal{N}:
M = B cup N,
end{equation*}
where $mathcal{N}$ is the set of borel-null-sets.
Now define
begin{equation*}
mathcal{D}
:= { B in mathcal{B}(mathbb{R}^d): lambda(B) = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset B } }
end{equation*}
as the set of all inner regular sets in the borel $sigma$-algebra, which, by construction is a subset of $mathcal{B}(mathbb{R}^d)$.
Now we want to show, that $mathcal{B}(mathbb{R}^d) subset mathcal{D}$ to conclude $mathcal{B}(mathbb{R}^d) = mathcal{D}$.
From the above lemma we know, that for all open sets $mathcal{O} subset mathbb{R}^d$ we have $mathcal{O} in mathcal{D}$, since by the continuity of the Lebesgue-measure, for all $varepsilon > 0$ we have $lambda(A) le lambda(bigcup_{n in mathbb{N}} W_n) + varepsilon$.
Since the set of open sets is a $cap$-stable generator of $mathcal{B}(mathbb{R}^d) subset mathcal{P}(mathbb{R})^d$, we only need to show that $mathcal{D}$ is a dynkin system. (German Wikipedia article on this line of argument)
We have $mathbb{R}^d in mathcal{D}$, because $lambda(mathbb{R}^d) = infty = sup{lambda(K): K subset mathbb{R}^d }$.
Let $D in mathcal{D}$. Then ??
Let ${ A_n }_{n in mathbb{N}} subset mathcal{D}$ be a family of disjoint subsets. Then
real-analysis measure-theory lebesgue-measure
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
$endgroup$
add a comment |
$begingroup$
In a previous homework our task was to prove the regularity of the Lebegue-measure on $mathbb{R}^d$.
More precisely:
Let $(mathbb{R}^d, mathcal{M}^*, lambda)$ be a measure space and $mathcal{M}^* := mathcal{M}^*(mathbb{R}^d)$ be the $sigma$-algebra of the $lambda^*$-measurable sets, with $lambda$ being the by the Lebesgue-outer-measure $lambda^*$ induced measure.
Show that
begin{align*}
lambda(A)
& = inf{lambda(O) mid O subset mathbb{R}^d text{ is open and } A subset O } \
& = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset A }.
end{align*}
Our attempt goes as follows:
For all $A in mathcal{M}^*$ we want to show
begin{equation*}
lambda(A)
ge inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
ge sup{ lambda(K): K subset mathbb{R}^d text{ compact and } K subset A }
ge lambda(A)
end{equation*}
First inequality:
If $lambda(A) = infty$, because of the monotonicity of the measure, we have $lambda(O) = infty$ for all open sets $O$ with $A subset O$.
If $lambda(A) < infty$, we define $A_{delta} := bigcup_{a in A} U_{delta}(a)$, which is a open and therefore measurable set with $A subset A_{delta}$.
Now we have $A_{delta} setminus A xrightarrow{delta to 0} emptyset$ and because of the continuity from below in the emptyset of $lambda$, and, because $A_{delta} setminus A$ is measurable, because it's the difference of measurable sets,
begin{equation*}
forall varepsilon > 0
exists delta > 0:
lambda(A_{delta} setminus A) < varepsilon.
end{equation*}
Because $A_{delta}$ is open and $sigma$-additivity of $lambda$ we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(A_{delta})
= lambda(A_{delta} setminus A) + lambda(A)
< varepsilon + lambda(A).
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
The second inequality follows from the monotonicity of $lambda^*$:
From $K subset A subset O$ we have $lambda(K) le lambda(O)$ for $K$ und $O$ as defined above.
Because taking the supremum or infimum doesn't change weak inequalities, the inequality follows.
The third inequality:
One can show, that every open set is $sigma$-compact.
Lemma
Let $A_{delta} subset mathbb{R}^d$ be an open subset.
Then there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$, so that $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Proof
Let $a in A_{delta} $.
Because $A_{delta} $ is open, there exists an $varepsilon_{a} > 0$, so that $U_{varepsilon_a}(a) subset A_{delta} $.
For every $a in A_{delta} $ we choose a bounded closed axially parallel cube $W_a$ with rational midpoint from $mathbb{Q}^d$ and rational edge length $q in mathbb{Q}$, so that
begin{equation*}
a
in W_{a}
subset U_{varepsilon_a}(a)
subset A_{delta} .
end{equation*}
This is always possible, because $mathbb{Q}$ is dense in $mathbb{R}$.
Therefore, we have $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Since there are only countable many cubes of this form, the union is countable. $square$
We let $K_n := bigcup_{j = 1}^{n} W_j$, which is compact as union of compact sets.
Then $K_n xrightarrow{n to infty} A_{delta}$.
Now we have $A_{delta} setminus K_n xrightarrow{n to infty} emptyset$ and with analogous argumentation as above
begin{equation*}
forall varepsilon > 0
exists N in mathbb{N}:
lambda(A_{delta} setminus K_n) < varepsilon
forall n > N.
end{equation*}
Therefore follows for all $varepsilon > 0$
begin{align*}
lambda(A)
le lambda(A_{delta})
le lambda(A_{delta}setminus K_n) + lambda(K_n)
< varepsilon + sup{ lambda(K): K subset mathbb{R}^d text{ compact und } K subset A_{delta} },
end{align*}
Because $varepsilon > 0$ was arbitrary and $delta$ can be arbitrarily small, the inequality follows.
My problem
I assume the proof of the second inequality is right. But: I not sure if the reasoning at the end of the proof of the last inequality is rigorous enough. Especially, if you take $A := mathbb{Q}$ in the proof of the first inequality, then for every $delta > 0$, we have $A_{delta} = mathbb{R}$ and therefore, we don't have $A_{delta} setminus A to emptyset$ or even $lambda(A_{delta} setminus A) to 0$.
Is there anyway to ''save'' this proof by fixing it and not changing the approach?
Correct Proof
First inequality
Case 1: $lambda(A) = infty$.
As above.
Case 2: $lambda(A) < infty$.
Utilising the Caratheodory construction of the Lebesgue measure, we know that
$$
forall varepsilon > 0
exists (a_n,b_n] := prod_{i=1}^d left(a_{n}^{(i)},b_{n}^{(i)}right]:
A subset bigcup_{n in mathbb{N}} (a_n,b_n]
quad text{and} quad
sum_{k=1}^infty lambda left((a_k,b_k]right)
< lambda(A) + frac{varepsilon}{2},
$$
where the $d$-dimensional cubes $(a_n,b_n]$ are pairwise disjoint.
Now let
$$
U := bigcup_{n=1}^infty (a_n, b_n+ t_n varepsilon)
qquad text{with} qquad t_n := 2^{-n-2d-1} max{1,b_{n}^{(i)} -a_{n}^{(i)}}^{-(d-1)}.
$$
Now we have $A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and $lambda(U) < lambda(A) + varepsilon$.
Therefore, we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(U)
< lambda(A) + varepsilon.
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
Second inequality
Same as above
Third inequality
From a previous homework we know that $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$ and therefore, that
begin{equation*}
forall M in mathcal{M}^*
exists B in mathcal{B}(mathbb{R}^d), N in mathcal{N}:
M = B cup N,
end{equation*}
where $mathcal{N}$ is the set of borel-null-sets.
Now define
begin{equation*}
mathcal{D}
:= { B in mathcal{B}(mathbb{R}^d): lambda(B) = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset B } }
end{equation*}
as the set of all inner regular sets in the borel $sigma$-algebra, which, by construction is a subset of $mathcal{B}(mathbb{R}^d)$.
Now we want to show, that $mathcal{B}(mathbb{R}^d) subset mathcal{D}$ to conclude $mathcal{B}(mathbb{R}^d) = mathcal{D}$.
From the above lemma we know, that for all open sets $mathcal{O} subset mathbb{R}^d$ we have $mathcal{O} in mathcal{D}$, since by the continuity of the Lebesgue-measure, for all $varepsilon > 0$ we have $lambda(A) le lambda(bigcup_{n in mathbb{N}} W_n) + varepsilon$.
Since the set of open sets is a $cap$-stable generator of $mathcal{B}(mathbb{R}^d) subset mathcal{P}(mathbb{R})^d$, we only need to show that $mathcal{D}$ is a dynkin system. (German Wikipedia article on this line of argument)
We have $mathbb{R}^d in mathcal{D}$, because $lambda(mathbb{R}^d) = infty = sup{lambda(K): K subset mathbb{R}^d }$.
Let $D in mathcal{D}$. Then ??
Let ${ A_n }_{n in mathbb{N}} subset mathcal{D}$ be a family of disjoint subsets. Then
real-analysis measure-theory lebesgue-measure
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
$endgroup$
add a comment |
$begingroup$
In a previous homework our task was to prove the regularity of the Lebegue-measure on $mathbb{R}^d$.
More precisely:
Let $(mathbb{R}^d, mathcal{M}^*, lambda)$ be a measure space and $mathcal{M}^* := mathcal{M}^*(mathbb{R}^d)$ be the $sigma$-algebra of the $lambda^*$-measurable sets, with $lambda$ being the by the Lebesgue-outer-measure $lambda^*$ induced measure.
Show that
begin{align*}
lambda(A)
& = inf{lambda(O) mid O subset mathbb{R}^d text{ is open and } A subset O } \
& = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset A }.
end{align*}
Our attempt goes as follows:
For all $A in mathcal{M}^*$ we want to show
begin{equation*}
lambda(A)
ge inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
ge sup{ lambda(K): K subset mathbb{R}^d text{ compact and } K subset A }
ge lambda(A)
end{equation*}
First inequality:
If $lambda(A) = infty$, because of the monotonicity of the measure, we have $lambda(O) = infty$ for all open sets $O$ with $A subset O$.
If $lambda(A) < infty$, we define $A_{delta} := bigcup_{a in A} U_{delta}(a)$, which is a open and therefore measurable set with $A subset A_{delta}$.
Now we have $A_{delta} setminus A xrightarrow{delta to 0} emptyset$ and because of the continuity from below in the emptyset of $lambda$, and, because $A_{delta} setminus A$ is measurable, because it's the difference of measurable sets,
begin{equation*}
forall varepsilon > 0
exists delta > 0:
lambda(A_{delta} setminus A) < varepsilon.
end{equation*}
Because $A_{delta}$ is open and $sigma$-additivity of $lambda$ we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(A_{delta})
= lambda(A_{delta} setminus A) + lambda(A)
< varepsilon + lambda(A).
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
The second inequality follows from the monotonicity of $lambda^*$:
From $K subset A subset O$ we have $lambda(K) le lambda(O)$ for $K$ und $O$ as defined above.
Because taking the supremum or infimum doesn't change weak inequalities, the inequality follows.
The third inequality:
One can show, that every open set is $sigma$-compact.
Lemma
Let $A_{delta} subset mathbb{R}^d$ be an open subset.
Then there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$, so that $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Proof
Let $a in A_{delta} $.
Because $A_{delta} $ is open, there exists an $varepsilon_{a} > 0$, so that $U_{varepsilon_a}(a) subset A_{delta} $.
For every $a in A_{delta} $ we choose a bounded closed axially parallel cube $W_a$ with rational midpoint from $mathbb{Q}^d$ and rational edge length $q in mathbb{Q}$, so that
begin{equation*}
a
in W_{a}
subset U_{varepsilon_a}(a)
subset A_{delta} .
end{equation*}
This is always possible, because $mathbb{Q}$ is dense in $mathbb{R}$.
Therefore, we have $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Since there are only countable many cubes of this form, the union is countable. $square$
We let $K_n := bigcup_{j = 1}^{n} W_j$, which is compact as union of compact sets.
Then $K_n xrightarrow{n to infty} A_{delta}$.
Now we have $A_{delta} setminus K_n xrightarrow{n to infty} emptyset$ and with analogous argumentation as above
begin{equation*}
forall varepsilon > 0
exists N in mathbb{N}:
lambda(A_{delta} setminus K_n) < varepsilon
forall n > N.
end{equation*}
Therefore follows for all $varepsilon > 0$
begin{align*}
lambda(A)
le lambda(A_{delta})
le lambda(A_{delta}setminus K_n) + lambda(K_n)
< varepsilon + sup{ lambda(K): K subset mathbb{R}^d text{ compact und } K subset A_{delta} },
end{align*}
Because $varepsilon > 0$ was arbitrary and $delta$ can be arbitrarily small, the inequality follows.
My problem
I assume the proof of the second inequality is right. But: I not sure if the reasoning at the end of the proof of the last inequality is rigorous enough. Especially, if you take $A := mathbb{Q}$ in the proof of the first inequality, then for every $delta > 0$, we have $A_{delta} = mathbb{R}$ and therefore, we don't have $A_{delta} setminus A to emptyset$ or even $lambda(A_{delta} setminus A) to 0$.
Is there anyway to ''save'' this proof by fixing it and not changing the approach?
Correct Proof
First inequality
Case 1: $lambda(A) = infty$.
As above.
Case 2: $lambda(A) < infty$.
Utilising the Caratheodory construction of the Lebesgue measure, we know that
$$
forall varepsilon > 0
exists (a_n,b_n] := prod_{i=1}^d left(a_{n}^{(i)},b_{n}^{(i)}right]:
A subset bigcup_{n in mathbb{N}} (a_n,b_n]
quad text{and} quad
sum_{k=1}^infty lambda left((a_k,b_k]right)
< lambda(A) + frac{varepsilon}{2},
$$
where the $d$-dimensional cubes $(a_n,b_n]$ are pairwise disjoint.
Now let
$$
U := bigcup_{n=1}^infty (a_n, b_n+ t_n varepsilon)
qquad text{with} qquad t_n := 2^{-n-2d-1} max{1,b_{n}^{(i)} -a_{n}^{(i)}}^{-(d-1)}.
$$
Now we have $A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and $lambda(U) < lambda(A) + varepsilon$.
Therefore, we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(U)
< lambda(A) + varepsilon.
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
Second inequality
Same as above
Third inequality
From a previous homework we know that $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$ and therefore, that
begin{equation*}
forall M in mathcal{M}^*
exists B in mathcal{B}(mathbb{R}^d), N in mathcal{N}:
M = B cup N,
end{equation*}
where $mathcal{N}$ is the set of borel-null-sets.
Now define
begin{equation*}
mathcal{D}
:= { B in mathcal{B}(mathbb{R}^d): lambda(B) = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset B } }
end{equation*}
as the set of all inner regular sets in the borel $sigma$-algebra, which, by construction is a subset of $mathcal{B}(mathbb{R}^d)$.
Now we want to show, that $mathcal{B}(mathbb{R}^d) subset mathcal{D}$ to conclude $mathcal{B}(mathbb{R}^d) = mathcal{D}$.
From the above lemma we know, that for all open sets $mathcal{O} subset mathbb{R}^d$ we have $mathcal{O} in mathcal{D}$, since by the continuity of the Lebesgue-measure, for all $varepsilon > 0$ we have $lambda(A) le lambda(bigcup_{n in mathbb{N}} W_n) + varepsilon$.
Since the set of open sets is a $cap$-stable generator of $mathcal{B}(mathbb{R}^d) subset mathcal{P}(mathbb{R})^d$, we only need to show that $mathcal{D}$ is a dynkin system. (German Wikipedia article on this line of argument)
We have $mathbb{R}^d in mathcal{D}$, because $lambda(mathbb{R}^d) = infty = sup{lambda(K): K subset mathbb{R}^d }$.
Let $D in mathcal{D}$. Then ??
Let ${ A_n }_{n in mathbb{N}} subset mathcal{D}$ be a family of disjoint subsets. Then
real-analysis measure-theory lebesgue-measure
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
$endgroup$
In a previous homework our task was to prove the regularity of the Lebegue-measure on $mathbb{R}^d$.
More precisely:
Let $(mathbb{R}^d, mathcal{M}^*, lambda)$ be a measure space and $mathcal{M}^* := mathcal{M}^*(mathbb{R}^d)$ be the $sigma$-algebra of the $lambda^*$-measurable sets, with $lambda$ being the by the Lebesgue-outer-measure $lambda^*$ induced measure.
Show that
begin{align*}
lambda(A)
& = inf{lambda(O) mid O subset mathbb{R}^d text{ is open and } A subset O } \
& = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset A }.
end{align*}
Our attempt goes as follows:
For all $A in mathcal{M}^*$ we want to show
begin{equation*}
lambda(A)
ge inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
ge sup{ lambda(K): K subset mathbb{R}^d text{ compact and } K subset A }
ge lambda(A)
end{equation*}
First inequality:
If $lambda(A) = infty$, because of the monotonicity of the measure, we have $lambda(O) = infty$ for all open sets $O$ with $A subset O$.
If $lambda(A) < infty$, we define $A_{delta} := bigcup_{a in A} U_{delta}(a)$, which is a open and therefore measurable set with $A subset A_{delta}$.
Now we have $A_{delta} setminus A xrightarrow{delta to 0} emptyset$ and because of the continuity from below in the emptyset of $lambda$, and, because $A_{delta} setminus A$ is measurable, because it's the difference of measurable sets,
begin{equation*}
forall varepsilon > 0
exists delta > 0:
lambda(A_{delta} setminus A) < varepsilon.
end{equation*}
Because $A_{delta}$ is open and $sigma$-additivity of $lambda$ we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(A_{delta})
= lambda(A_{delta} setminus A) + lambda(A)
< varepsilon + lambda(A).
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
The second inequality follows from the monotonicity of $lambda^*$:
From $K subset A subset O$ we have $lambda(K) le lambda(O)$ for $K$ und $O$ as defined above.
Because taking the supremum or infimum doesn't change weak inequalities, the inequality follows.
The third inequality:
One can show, that every open set is $sigma$-compact.
Lemma
Let $A_{delta} subset mathbb{R}^d$ be an open subset.
Then there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$, so that $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Proof
Let $a in A_{delta} $.
Because $A_{delta} $ is open, there exists an $varepsilon_{a} > 0$, so that $U_{varepsilon_a}(a) subset A_{delta} $.
For every $a in A_{delta} $ we choose a bounded closed axially parallel cube $W_a$ with rational midpoint from $mathbb{Q}^d$ and rational edge length $q in mathbb{Q}$, so that
begin{equation*}
a
in W_{a}
subset U_{varepsilon_a}(a)
subset A_{delta} .
end{equation*}
This is always possible, because $mathbb{Q}$ is dense in $mathbb{R}$.
Therefore, we have $A_{delta} = bigcup_{n in mathbb{N}} W_n$.
Since there are only countable many cubes of this form, the union is countable. $square$
We let $K_n := bigcup_{j = 1}^{n} W_j$, which is compact as union of compact sets.
Then $K_n xrightarrow{n to infty} A_{delta}$.
Now we have $A_{delta} setminus K_n xrightarrow{n to infty} emptyset$ and with analogous argumentation as above
begin{equation*}
forall varepsilon > 0
exists N in mathbb{N}:
lambda(A_{delta} setminus K_n) < varepsilon
forall n > N.
end{equation*}
Therefore follows for all $varepsilon > 0$
begin{align*}
lambda(A)
le lambda(A_{delta})
le lambda(A_{delta}setminus K_n) + lambda(K_n)
< varepsilon + sup{ lambda(K): K subset mathbb{R}^d text{ compact und } K subset A_{delta} },
end{align*}
Because $varepsilon > 0$ was arbitrary and $delta$ can be arbitrarily small, the inequality follows.
My problem
I assume the proof of the second inequality is right. But: I not sure if the reasoning at the end of the proof of the last inequality is rigorous enough. Especially, if you take $A := mathbb{Q}$ in the proof of the first inequality, then for every $delta > 0$, we have $A_{delta} = mathbb{R}$ and therefore, we don't have $A_{delta} setminus A to emptyset$ or even $lambda(A_{delta} setminus A) to 0$.
Is there anyway to ''save'' this proof by fixing it and not changing the approach?
Correct Proof
First inequality
Case 1: $lambda(A) = infty$.
As above.
Case 2: $lambda(A) < infty$.
Utilising the Caratheodory construction of the Lebesgue measure, we know that
$$
forall varepsilon > 0
exists (a_n,b_n] := prod_{i=1}^d left(a_{n}^{(i)},b_{n}^{(i)}right]:
A subset bigcup_{n in mathbb{N}} (a_n,b_n]
quad text{and} quad
sum_{k=1}^infty lambda left((a_k,b_k]right)
< lambda(A) + frac{varepsilon}{2},
$$
where the $d$-dimensional cubes $(a_n,b_n]$ are pairwise disjoint.
Now let
$$
U := bigcup_{n=1}^infty (a_n, b_n+ t_n varepsilon)
qquad text{with} qquad t_n := 2^{-n-2d-1} max{1,b_{n}^{(i)} -a_{n}^{(i)}}^{-(d-1)}.
$$
Now we have $A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and $lambda(U) < lambda(A) + varepsilon$.
Therefore, we have
begin{equation*}
inf{ lambda(O): O subset mathbb{R}^d text{ open und } A subset O }
le lambda(U)
< lambda(A) + varepsilon.
end{equation*}
Because $varepsilon > 0$ was arbitrary, the inequality follows.
Second inequality
Same as above
Third inequality
From a previous homework we know that $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$ and therefore, that
begin{equation*}
forall M in mathcal{M}^*
exists B in mathcal{B}(mathbb{R}^d), N in mathcal{N}:
M = B cup N,
end{equation*}
where $mathcal{N}$ is the set of borel-null-sets.
Now define
begin{equation*}
mathcal{D}
:= { B in mathcal{B}(mathbb{R}^d): lambda(B) = sup{lambda(K) mid K subset mathbb{R}^d text{ is compact and } K subset B } }
end{equation*}
as the set of all inner regular sets in the borel $sigma$-algebra, which, by construction is a subset of $mathcal{B}(mathbb{R}^d)$.
Now we want to show, that $mathcal{B}(mathbb{R}^d) subset mathcal{D}$ to conclude $mathcal{B}(mathbb{R}^d) = mathcal{D}$.
From the above lemma we know, that for all open sets $mathcal{O} subset mathbb{R}^d$ we have $mathcal{O} in mathcal{D}$, since by the continuity of the Lebesgue-measure, for all $varepsilon > 0$ we have $lambda(A) le lambda(bigcup_{n in mathbb{N}} W_n) + varepsilon$.
Since the set of open sets is a $cap$-stable generator of $mathcal{B}(mathbb{R}^d) subset mathcal{P}(mathbb{R})^d$, we only need to show that $mathcal{D}$ is a dynkin system. (German Wikipedia article on this line of argument)
We have $mathbb{R}^d in mathcal{D}$, because $lambda(mathbb{R}^d) = infty = sup{lambda(K): K subset mathbb{R}^d }$.
Let $D in mathcal{D}$. Then ??
Let ${ A_n }_{n in mathbb{N}} subset mathcal{D}$ be a family of disjoint subsets. Then
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
edited Jan 13 at 20:50
Viktor Glombik
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
asked Jan 10 at 16:56
Viktor GlombikViktor Glombik
8211527
8211527
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_delta = mathbb{R}$ and $K_n$ will approximate the measure of $mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $lambda(A) < infty$, then we can find $(a_n,b_n] = prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$sum_{k=1}^infty lambda ((a_n,b_n]) < lambda(A) + varepsilon/2$$
Thus, we can take $U= bigcup_{n=1}^infty (a_n,b_n+ t_n varepsilon)$. (Note $ A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} max{1,b_{n,i} -a_{n,i}}^{-(d-1)} $ we get $lambda(U) < lambda(A) + varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $mathcal{M}^*$ is the completion of the Borel-$sigma$-algebra. Thus any $A in mathcal{M}^*$ can be written as $A= B cup M$ with a Borel set $B$ and $M subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$mathcal{D} := { B in mathcal{B}(mathbb{R}^d) : B text{ is inner regular}}.$$
- Your argument shows that open sets are in $mathcal{D}$.
- Check that $mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $cap$-stable generator of the Borel-$sigma$-algebra we can conclude that $mathcal{D} = mathcal{B}(mathbb{R}^d)$.
I additionally added a prove of 2: Note that $mathbb{R}^d$ is an open set. If $(A_n)_{in mathbb{N}} subset mathcal{D}$ are disjoint, then we can take compact $K_n subset A_n$ with $lambda(A_n) < lambda(K_n) + varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $lambda(cup_{n=1}^infty A_n) = sum_{n=1}^infty lambda(A_n) = infty$, then for any $K>0$ there exists $N$ such that $$sum_{n=1}^N lambda(A_n) >K.$$ Thus $sum_{n=1}^N lambda(K_n) > K-varepsilon$. So taking the compact set $K = cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $infty$.
In the second case, we have for some $N in mathbb{N}$, because the series is convergent, that $$lambda(cup_{n=1}^infty A_n) < sum_{n=1}^N lambda(A_n) + varepsilon.$$
Thus
$$lambda(cup_{n=1}^infty A_n) < 2 varepsilon + lambda(cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = cup_{n=1}^N K_n$. This proves that $cup_{n=1}^infty A_n in mathcal{D}$.
Let $A,B in mathcal{D}$ with $B subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A cap prod_{i=1}^d (n_i,n_i+1] quad text{and} quad B_i = B cap prod_{i=1}^d (n_i,n_i+1]$$ with $n_i in mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i setminus B_i in mathcal{D}$. then also the union of this disjoint sets is in $mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i in mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B subset U$ and $lambda(U setminus B) < varepsilon$ (this is possible, because we already know that $lambda$ is outer regular) and a compact set with $K subset A$ and $lambda(A setminus K) < varepsilon$. Define $L = K setminus U$. Then $L$ is compact and
$$lambda( (A setminus B) setminus L) le lambda(U setminus B) + lambda(A setminus K) < 2 varepsilon.$$
Therefore $A setminus B in mathcal{D}$.
answered Jan 13 at 12:20
p4schp4sch
5,275217
$endgroup$
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
|
show 6 more comments
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1 Answer
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$begingroup$
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_delta = mathbb{R}$ and $K_n$ will approximate the measure of $mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $lambda(A) < infty$, then we can find $(a_n,b_n] = prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$sum_{k=1}^infty lambda ((a_n,b_n]) < lambda(A) + varepsilon/2$$
Thus, we can take $U= bigcup_{n=1}^infty (a_n,b_n+ t_n varepsilon)$. (Note $ A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} max{1,b_{n,i} -a_{n,i}}^{-(d-1)} $ we get $lambda(U) < lambda(A) + varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $mathcal{M}^*$ is the completion of the Borel-$sigma$-algebra. Thus any $A in mathcal{M}^*$ can be written as $A= B cup M$ with a Borel set $B$ and $M subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$mathcal{D} := { B in mathcal{B}(mathbb{R}^d) : B text{ is inner regular}}.$$
- Your argument shows that open sets are in $mathcal{D}$.
- Check that $mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $cap$-stable generator of the Borel-$sigma$-algebra we can conclude that $mathcal{D} = mathcal{B}(mathbb{R}^d)$.
I additionally added a prove of 2: Note that $mathbb{R}^d$ is an open set. If $(A_n)_{in mathbb{N}} subset mathcal{D}$ are disjoint, then we can take compact $K_n subset A_n$ with $lambda(A_n) < lambda(K_n) + varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $lambda(cup_{n=1}^infty A_n) = sum_{n=1}^infty lambda(A_n) = infty$, then for any $K>0$ there exists $N$ such that $$sum_{n=1}^N lambda(A_n) >K.$$ Thus $sum_{n=1}^N lambda(K_n) > K-varepsilon$. So taking the compact set $K = cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $infty$.
In the second case, we have for some $N in mathbb{N}$, because the series is convergent, that $$lambda(cup_{n=1}^infty A_n) < sum_{n=1}^N lambda(A_n) + varepsilon.$$
Thus
$$lambda(cup_{n=1}^infty A_n) < 2 varepsilon + lambda(cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = cup_{n=1}^N K_n$. This proves that $cup_{n=1}^infty A_n in mathcal{D}$.
Let $A,B in mathcal{D}$ with $B subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A cap prod_{i=1}^d (n_i,n_i+1] quad text{and} quad B_i = B cap prod_{i=1}^d (n_i,n_i+1]$$ with $n_i in mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i setminus B_i in mathcal{D}$. then also the union of this disjoint sets is in $mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i in mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B subset U$ and $lambda(U setminus B) < varepsilon$ (this is possible, because we already know that $lambda$ is outer regular) and a compact set with $K subset A$ and $lambda(A setminus K) < varepsilon$. Define $L = K setminus U$. Then $L$ is compact and
$$lambda( (A setminus B) setminus L) le lambda(U setminus B) + lambda(A setminus K) < 2 varepsilon.$$
Therefore $A setminus B in mathcal{D}$.
answered Jan 13 at 12:20
p4schp4sch
5,275217
$endgroup$
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
|
show 6 more comments
$begingroup$
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_delta = mathbb{R}$ and $K_n$ will approximate the measure of $mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $lambda(A) < infty$, then we can find $(a_n,b_n] = prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$sum_{k=1}^infty lambda ((a_n,b_n]) < lambda(A) + varepsilon/2$$
Thus, we can take $U= bigcup_{n=1}^infty (a_n,b_n+ t_n varepsilon)$. (Note $ A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} max{1,b_{n,i} -a_{n,i}}^{-(d-1)} $ we get $lambda(U) < lambda(A) + varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $mathcal{M}^*$ is the completion of the Borel-$sigma$-algebra. Thus any $A in mathcal{M}^*$ can be written as $A= B cup M$ with a Borel set $B$ and $M subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$mathcal{D} := { B in mathcal{B}(mathbb{R}^d) : B text{ is inner regular}}.$$
- Your argument shows that open sets are in $mathcal{D}$.
- Check that $mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $cap$-stable generator of the Borel-$sigma$-algebra we can conclude that $mathcal{D} = mathcal{B}(mathbb{R}^d)$.
I additionally added a prove of 2: Note that $mathbb{R}^d$ is an open set. If $(A_n)_{in mathbb{N}} subset mathcal{D}$ are disjoint, then we can take compact $K_n subset A_n$ with $lambda(A_n) < lambda(K_n) + varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $lambda(cup_{n=1}^infty A_n) = sum_{n=1}^infty lambda(A_n) = infty$, then for any $K>0$ there exists $N$ such that $$sum_{n=1}^N lambda(A_n) >K.$$ Thus $sum_{n=1}^N lambda(K_n) > K-varepsilon$. So taking the compact set $K = cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $infty$.
In the second case, we have for some $N in mathbb{N}$, because the series is convergent, that $$lambda(cup_{n=1}^infty A_n) < sum_{n=1}^N lambda(A_n) + varepsilon.$$
Thus
$$lambda(cup_{n=1}^infty A_n) < 2 varepsilon + lambda(cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = cup_{n=1}^N K_n$. This proves that $cup_{n=1}^infty A_n in mathcal{D}$.
Let $A,B in mathcal{D}$ with $B subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A cap prod_{i=1}^d (n_i,n_i+1] quad text{and} quad B_i = B cap prod_{i=1}^d (n_i,n_i+1]$$ with $n_i in mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i setminus B_i in mathcal{D}$. then also the union of this disjoint sets is in $mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i in mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B subset U$ and $lambda(U setminus B) < varepsilon$ (this is possible, because we already know that $lambda$ is outer regular) and a compact set with $K subset A$ and $lambda(A setminus K) < varepsilon$. Define $L = K setminus U$. Then $L$ is compact and
$$lambda( (A setminus B) setminus L) le lambda(U setminus B) + lambda(A setminus K) < 2 varepsilon.$$
Therefore $A setminus B in mathcal{D}$.
answered Jan 13 at 12:20
p4schp4sch
5,275217
$endgroup$
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
|
show 6 more comments
$begingroup$
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_delta = mathbb{R}$ and $K_n$ will approximate the measure of $mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $lambda(A) < infty$, then we can find $(a_n,b_n] = prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$sum_{k=1}^infty lambda ((a_n,b_n]) < lambda(A) + varepsilon/2$$
Thus, we can take $U= bigcup_{n=1}^infty (a_n,b_n+ t_n varepsilon)$. (Note $ A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} max{1,b_{n,i} -a_{n,i}}^{-(d-1)} $ we get $lambda(U) < lambda(A) + varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $mathcal{M}^*$ is the completion of the Borel-$sigma$-algebra. Thus any $A in mathcal{M}^*$ can be written as $A= B cup M$ with a Borel set $B$ and $M subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$mathcal{D} := { B in mathcal{B}(mathbb{R}^d) : B text{ is inner regular}}.$$
- Your argument shows that open sets are in $mathcal{D}$.
- Check that $mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $cap$-stable generator of the Borel-$sigma$-algebra we can conclude that $mathcal{D} = mathcal{B}(mathbb{R}^d)$.
I additionally added a prove of 2: Note that $mathbb{R}^d$ is an open set. If $(A_n)_{in mathbb{N}} subset mathcal{D}$ are disjoint, then we can take compact $K_n subset A_n$ with $lambda(A_n) < lambda(K_n) + varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $lambda(cup_{n=1}^infty A_n) = sum_{n=1}^infty lambda(A_n) = infty$, then for any $K>0$ there exists $N$ such that $$sum_{n=1}^N lambda(A_n) >K.$$ Thus $sum_{n=1}^N lambda(K_n) > K-varepsilon$. So taking the compact set $K = cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $infty$.
In the second case, we have for some $N in mathbb{N}$, because the series is convergent, that $$lambda(cup_{n=1}^infty A_n) < sum_{n=1}^N lambda(A_n) + varepsilon.$$
Thus
$$lambda(cup_{n=1}^infty A_n) < 2 varepsilon + lambda(cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = cup_{n=1}^N K_n$. This proves that $cup_{n=1}^infty A_n in mathcal{D}$.
Let $A,B in mathcal{D}$ with $B subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A cap prod_{i=1}^d (n_i,n_i+1] quad text{and} quad B_i = B cap prod_{i=1}^d (n_i,n_i+1]$$ with $n_i in mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i setminus B_i in mathcal{D}$. then also the union of this disjoint sets is in $mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i in mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B subset U$ and $lambda(U setminus B) < varepsilon$ (this is possible, because we already know that $lambda$ is outer regular) and a compact set with $K subset A$ and $lambda(A setminus K) < varepsilon$. Define $L = K setminus U$. Then $L$ is compact and
$$lambda( (A setminus B) setminus L) le lambda(U setminus B) + lambda(A setminus K) < 2 varepsilon.$$
Therefore $A setminus B in mathcal{D}$.
answered Jan 13 at 12:20
p4schp4sch
5,275217
$endgroup$
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_delta = mathbb{R}$ and $K_n$ will approximate the measure of $mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $lambda(A) < infty$, then we can find $(a_n,b_n] = prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$sum_{k=1}^infty lambda ((a_n,b_n]) < lambda(A) + varepsilon/2$$
Thus, we can take $U= bigcup_{n=1}^infty (a_n,b_n+ t_n varepsilon)$. (Note $ A subset bigcup_{n=1}^infty (a_n,b_n] subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} max{1,b_{n,i} -a_{n,i}}^{-(d-1)} $ we get $lambda(U) < lambda(A) + varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $mathcal{M}^*$ is the completion of the Borel-$sigma$-algebra. Thus any $A in mathcal{M}^*$ can be written as $A= B cup M$ with a Borel set $B$ and $M subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$mathcal{D} := { B in mathcal{B}(mathbb{R}^d) : B text{ is inner regular}}.$$
- Your argument shows that open sets are in $mathcal{D}$.
- Check that $mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $cap$-stable generator of the Borel-$sigma$-algebra we can conclude that $mathcal{D} = mathcal{B}(mathbb{R}^d)$.
I additionally added a prove of 2: Note that $mathbb{R}^d$ is an open set. If $(A_n)_{in mathbb{N}} subset mathcal{D}$ are disjoint, then we can take compact $K_n subset A_n$ with $lambda(A_n) < lambda(K_n) + varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $lambda(cup_{n=1}^infty A_n) = sum_{n=1}^infty lambda(A_n) = infty$, then for any $K>0$ there exists $N$ such that $$sum_{n=1}^N lambda(A_n) >K.$$ Thus $sum_{n=1}^N lambda(K_n) > K-varepsilon$. So taking the compact set $K = cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $infty$.
In the second case, we have for some $N in mathbb{N}$, because the series is convergent, that $$lambda(cup_{n=1}^infty A_n) < sum_{n=1}^N lambda(A_n) + varepsilon.$$
Thus
$$lambda(cup_{n=1}^infty A_n) < 2 varepsilon + lambda(cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = cup_{n=1}^N K_n$. This proves that $cup_{n=1}^infty A_n in mathcal{D}$.
Let $A,B in mathcal{D}$ with $B subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A cap prod_{i=1}^d (n_i,n_i+1] quad text{and} quad B_i = B cap prod_{i=1}^d (n_i,n_i+1]$$ with $n_i in mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i setminus B_i in mathcal{D}$. then also the union of this disjoint sets is in $mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i in mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B subset U$ and $lambda(U setminus B) < varepsilon$ (this is possible, because we already know that $lambda$ is outer regular) and a compact set with $K subset A$ and $lambda(A setminus K) < varepsilon$. Define $L = K setminus U$. Then $L$ is compact and
$$lambda( (A setminus B) setminus L) le lambda(U setminus B) + lambda(A setminus K) < 2 varepsilon.$$
Therefore $A setminus B in mathcal{D}$.
answered Jan 13 at 12:20
p4schp4sch
5,275217
edited Jan 13 at 21:12
answered Jan 13 at 12:20
p4schp4sch
5,275217
answered Jan 13 at 12:20
p4schp4sch
5,275217
answered Jan 13 at 12:20
p4schp4sch
5,275217
5,275217
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
|
show 6 more comments
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
$begingroup$
For the first inequality, we had the idea that we can instead choose $A_{delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $varepsilon > 0$. Would this work?
$endgroup$
– Viktor Glombik
Jan 13 at 13:48
1
1
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = bigcup_{n=1}^infty (q_n -2^{-n} varepsilon, q_n +2^{-n} varepsilon)$, where $(q_n)_n$ is an enumeration of $mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset.
$endgroup$
– p4sch
Jan 13 at 13:54
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
$begingroup$
I have added the construction of the $W_j$ with the proof of the $sigma$-compactness of the open set $A_{delta}$. Can you please explain why it is a problem, that $K_n notsubset A$?
$endgroup$
– Viktor Glombik
Jan 13 at 15:05
1
1
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
$begingroup$
The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k in mathbb{N}}$ such that $A=bigcup_{n in mathbb{N}} W_n$. Thus, by the continuity of measures, we have $lambda(U) < lambda(bigcup_{n=1}^N W_n) + varepsilon$ for some $N in mathbb{N}$. Note that $K:=bigcup_{n=1}^N W_n$ is a compact set.
$endgroup$
– p4sch
Jan 13 at 20:40
1
1
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
$begingroup$
Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M subset N$) is not just a consequence of $mathcal{B}(mathbb{R}^d) subset mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space.
$endgroup$
– p4sch
Jan 13 at 21:23
|
show 6 more comments
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