Mixing
Question regarding a property of the Lebesgue measure on Wikipedia
$begingroup$
On wikipedia, the following property of the Lebesgue measure is listed:
If $A$ is a disjoint union of countably many disjoint Lebesgue-measurable sets, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
which I have interpreted as saying that
If $lbrace mathcal{X}_i:iin Irbrace $ is a family of countably many sets, then let $A$ satisfy
$$
A=bigsqcup_{iin I}mathcal{X}_i
$$
If $mathcal{X}_i$ and $mathcal{X}_j$ are disjoint for any $i,jin I$, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
Now, I am not certain if this interpretation is correct or not, so any corrections are appreciated, but above all else I find the fact that $A$ has to be the disjoint union rather redundant, since, if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $ineq j$, then $mu(mathcal{X}_icupmathcal{X}_j)=mu(mathcal{X}_isqcupmathcal{X}_j)$, so why bother with a disjoint union instead of a standard union? Is there a flaw in my logic?
measure-theory
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
add a comment |
$begingroup$
On wikipedia, the following property of the Lebesgue measure is listed:
If $A$ is a disjoint union of countably many disjoint Lebesgue-measurable sets, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
which I have interpreted as saying that
If $lbrace mathcal{X}_i:iin Irbrace $ is a family of countably many sets, then let $A$ satisfy
$$
A=bigsqcup_{iin I}mathcal{X}_i
$$
If $mathcal{X}_i$ and $mathcal{X}_j$ are disjoint for any $i,jin I$, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
Now, I am not certain if this interpretation is correct or not, so any corrections are appreciated, but above all else I find the fact that $A$ has to be the disjoint union rather redundant, since, if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $ineq j$, then $mu(mathcal{X}_icupmathcal{X}_j)=mu(mathcal{X}_isqcupmathcal{X}_j)$, so why bother with a disjoint union instead of a standard union? Is there a flaw in my logic?
measure-theory
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21
add a comment |
$begingroup$
On wikipedia, the following property of the Lebesgue measure is listed:
If $A$ is a disjoint union of countably many disjoint Lebesgue-measurable sets, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
which I have interpreted as saying that
If $lbrace mathcal{X}_i:iin Irbrace $ is a family of countably many sets, then let $A$ satisfy
$$
A=bigsqcup_{iin I}mathcal{X}_i
$$
If $mathcal{X}_i$ and $mathcal{X}_j$ are disjoint for any $i,jin I$, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
Now, I am not certain if this interpretation is correct or not, so any corrections are appreciated, but above all else I find the fact that $A$ has to be the disjoint union rather redundant, since, if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $ineq j$, then $mu(mathcal{X}_icupmathcal{X}_j)=mu(mathcal{X}_isqcupmathcal{X}_j)$, so why bother with a disjoint union instead of a standard union? Is there a flaw in my logic?
measure-theory
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
On wikipedia, the following property of the Lebesgue measure is listed:
If $A$ is a disjoint union of countably many disjoint Lebesgue-measurable sets, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
which I have interpreted as saying that
If $lbrace mathcal{X}_i:iin Irbrace $ is a family of countably many sets, then let $A$ satisfy
$$
A=bigsqcup_{iin I}mathcal{X}_i
$$
If $mathcal{X}_i$ and $mathcal{X}_j$ are disjoint for any $i,jin I$, then $A$ is itself Lebesgue-measurable and $mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets
Now, I am not certain if this interpretation is correct or not, so any corrections are appreciated, but above all else I find the fact that $A$ has to be the disjoint union rather redundant, since, if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $ineq j$, then $mu(mathcal{X}_icupmathcal{X}_j)=mu(mathcal{X}_isqcupmathcal{X}_j)$, so why bother with a disjoint union instead of a standard union? Is there a flaw in my logic?
measure-theory
measure-theory
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
edited Jan 14 at 4:01
Andrés E. Caicedo
65.4k8158249
65.4k8158249
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 13 at 18:16
joshuaheckroodtjoshuaheckroodt
1,205622
1,205622
$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21
add a comment |
$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21
$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21
$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.
Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
$endgroup$
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
add a comment |
$begingroup$
If $A_n, n in mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = bigcup_{n in mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $sigma$-algebra.
If moreover the sets $A_n$ obey $A_n cap A_m = emptyset$ for $n neq m$ then an axiom that by definition must hold for all measures tells us that:
$$mu(bigcup_{n in mathbb{N}} A_n) =sum_{n in mathbb{N}} mu(A_n)$$
and this does need disjointness.
For normal unions we can merely say:
$$mu(bigcup_{n in mathbb{N}} A_n) le sum_{n in mathbb{N}} mu(A_n)$$
the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $sigma$-algebra manipulation).
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.
Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
$endgroup$
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
add a comment |
$begingroup$
The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.
Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
$endgroup$
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
add a comment |
$begingroup$
The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.
Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
$endgroup$
The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.
Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
answered Jan 13 at 18:23
jmerryjmerry
8,6931123
8,6931123
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
add a comment |
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
So if every $mathcal{X}_i$ is disjoint from $mathcal{X}_j$ for $jneq i$, then $A$ doesn't have to be the disjoint union of all $mathcal{X}_i$, it can be merely a standard union?
$endgroup$
– joshuaheckroodt
Jan 13 at 18:30
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
$begingroup$
In practice, what you'll actually use is pretty much always an ordinary union of disjoint sets. Using a disjoint union would require you to extend the measure to a new space.
$endgroup$
– jmerry
Jan 13 at 18:35
add a comment |
$begingroup$
If $A_n, n in mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = bigcup_{n in mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $sigma$-algebra.
If moreover the sets $A_n$ obey $A_n cap A_m = emptyset$ for $n neq m$ then an axiom that by definition must hold for all measures tells us that:
$$mu(bigcup_{n in mathbb{N}} A_n) =sum_{n in mathbb{N}} mu(A_n)$$
and this does need disjointness.
For normal unions we can merely say:
$$mu(bigcup_{n in mathbb{N}} A_n) le sum_{n in mathbb{N}} mu(A_n)$$
the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $sigma$-algebra manipulation).
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
add a comment |
$begingroup$
If $A_n, n in mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = bigcup_{n in mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $sigma$-algebra.
If moreover the sets $A_n$ obey $A_n cap A_m = emptyset$ for $n neq m$ then an axiom that by definition must hold for all measures tells us that:
$$mu(bigcup_{n in mathbb{N}} A_n) =sum_{n in mathbb{N}} mu(A_n)$$
and this does need disjointness.
For normal unions we can merely say:
$$mu(bigcup_{n in mathbb{N}} A_n) le sum_{n in mathbb{N}} mu(A_n)$$
the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $sigma$-algebra manipulation).
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
add a comment |
$begingroup$
If $A_n, n in mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = bigcup_{n in mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $sigma$-algebra.
If moreover the sets $A_n$ obey $A_n cap A_m = emptyset$ for $n neq m$ then an axiom that by definition must hold for all measures tells us that:
$$mu(bigcup_{n in mathbb{N}} A_n) =sum_{n in mathbb{N}} mu(A_n)$$
and this does need disjointness.
For normal unions we can merely say:
$$mu(bigcup_{n in mathbb{N}} A_n) le sum_{n in mathbb{N}} mu(A_n)$$
the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $sigma$-algebra manipulation).
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
If $A_n, n in mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = bigcup_{n in mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $sigma$-algebra.
If moreover the sets $A_n$ obey $A_n cap A_m = emptyset$ for $n neq m$ then an axiom that by definition must hold for all measures tells us that:
$$mu(bigcup_{n in mathbb{N}} A_n) =sum_{n in mathbb{N}} mu(A_n)$$
and this does need disjointness.
For normal unions we can merely say:
$$mu(bigcup_{n in mathbb{N}} A_n) le sum_{n in mathbb{N}} mu(A_n)$$
the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $sigma$-algebra manipulation).
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 13 at 22:50
Henno BrandsmaHenno Brandsma
109k347115
109k347115
add a comment |
add a comment |
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$begingroup$
The statement is that if $X$ and $Y$ are disjoint, the $mu(Xcup Y)=mu(X)+mu(Y)$
$endgroup$
– saulspatz
Jan 13 at 18:21