Mixing
$f$ is $mathcal{F} - mathcal{B}$ measurable $iff$ $A in sigma(mathcal{F} times mathcal{B})$
$begingroup$
Let $(Omega,mathcal{F})$ be a measure space and $f: Omega rightarrow [0,infty]$ a nonnegative function and let
$A$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f(x)$}. Show:
$f$ is $mathcal{F} - mathcal{B}$ measurable $iff$ $A in sigma(mathcal{F} times mathcal{B})$
Where $mathcal{B}$ denotes the Borel-$sigma$-Algebra generated from the open sets on $[0,infty)$ and {$infty$}.
I think I am going in the correct direction of one implication:
Let $f$ be measurable $rightarrow$ $f$ is a nonnegative measurable function, we have shown there exist simple function $f_n$ with $f_n rightarrow f $ for $n rightarrow infty$ where $f_n =sum_{k=1}^{n2^n} (k-1)2^{-n} mathcal{1}_{(k-1)2^{-n}le f < k2^{-n}} + nmathcal{1}_{n le f}$ with $mathcal{1}$ being the indicator-function.
Define
$A_n$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f_n(x)$}= $bigcup_{k=1}^{n*2^n}$ {$fin[(k-1)2^{-n},k2^{-n}),yin[0,(k-1)2^{-n})$}. $A$ is the union of those $A_n$. Is it correct, that the set where f is in is the inverse image of an interval under a measurable function, hence measurable and the set in which y is an element of is in $mathcal{B}$, which would imply $A_n$ in $sigma(mathcal{F} times mathcal{B})$, so ultimately $A$ because it is the countable union of those $A_n$.
For the other implication I could not come up with anything good so far. Because I do not really know what $mathcal{F}$ is, I will have to use
$f$ is $mathcal{F} - mathcal{B}$ measurable if $f^{-1}(mathcal{B})subset mathcal{F}$.
Maybe I am overlooking something, but how would one show this ?
measure-theory measurable-functions
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal{F})$ be a measure space and $f: Omega rightarrow [0,infty]$ a nonnegative function and let
$A$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f(x)$}. Show:
$f$ is $mathcal{F} - mathcal{B}$ measurable $iff$ $A in sigma(mathcal{F} times mathcal{B})$
Where $mathcal{B}$ denotes the Borel-$sigma$-Algebra generated from the open sets on $[0,infty)$ and {$infty$}.
I think I am going in the correct direction of one implication:
Let $f$ be measurable $rightarrow$ $f$ is a nonnegative measurable function, we have shown there exist simple function $f_n$ with $f_n rightarrow f $ for $n rightarrow infty$ where $f_n =sum_{k=1}^{n2^n} (k-1)2^{-n} mathcal{1}_{(k-1)2^{-n}le f < k2^{-n}} + nmathcal{1}_{n le f}$ with $mathcal{1}$ being the indicator-function.
Define
$A_n$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f_n(x)$}= $bigcup_{k=1}^{n*2^n}$ {$fin[(k-1)2^{-n},k2^{-n}),yin[0,(k-1)2^{-n})$}. $A$ is the union of those $A_n$. Is it correct, that the set where f is in is the inverse image of an interval under a measurable function, hence measurable and the set in which y is an element of is in $mathcal{B}$, which would imply $A_n$ in $sigma(mathcal{F} times mathcal{B})$, so ultimately $A$ because it is the countable union of those $A_n$.
For the other implication I could not come up with anything good so far. Because I do not really know what $mathcal{F}$ is, I will have to use
$f$ is $mathcal{F} - mathcal{B}$ measurable if $f^{-1}(mathcal{B})subset mathcal{F}$.
Maybe I am overlooking something, but how would one show this ?
measure-theory measurable-functions
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal{F})$ be a measure space and $f: Omega rightarrow [0,infty]$ a nonnegative function and let
$A$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f(x)$}. Show:
$f$ is $mathcal{F} - mathcal{B}$ measurable $iff$ $A in sigma(mathcal{F} times mathcal{B})$
Where $mathcal{B}$ denotes the Borel-$sigma$-Algebra generated from the open sets on $[0,infty)$ and {$infty$}.
I think I am going in the correct direction of one implication:
Let $f$ be measurable $rightarrow$ $f$ is a nonnegative measurable function, we have shown there exist simple function $f_n$ with $f_n rightarrow f $ for $n rightarrow infty$ where $f_n =sum_{k=1}^{n2^n} (k-1)2^{-n} mathcal{1}_{(k-1)2^{-n}le f < k2^{-n}} + nmathcal{1}_{n le f}$ with $mathcal{1}$ being the indicator-function.
Define
$A_n$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f_n(x)$}= $bigcup_{k=1}^{n*2^n}$ {$fin[(k-1)2^{-n},k2^{-n}),yin[0,(k-1)2^{-n})$}. $A$ is the union of those $A_n$. Is it correct, that the set where f is in is the inverse image of an interval under a measurable function, hence measurable and the set in which y is an element of is in $mathcal{B}$, which would imply $A_n$ in $sigma(mathcal{F} times mathcal{B})$, so ultimately $A$ because it is the countable union of those $A_n$.
For the other implication I could not come up with anything good so far. Because I do not really know what $mathcal{F}$ is, I will have to use
$f$ is $mathcal{F} - mathcal{B}$ measurable if $f^{-1}(mathcal{B})subset mathcal{F}$.
Maybe I am overlooking something, but how would one show this ?
measure-theory measurable-functions
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
$endgroup$
Let $(Omega,mathcal{F})$ be a measure space and $f: Omega rightarrow [0,infty]$ a nonnegative function and let
$A$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f(x)$}. Show:
$f$ is $mathcal{F} - mathcal{B}$ measurable $iff$ $A in sigma(mathcal{F} times mathcal{B})$
Where $mathcal{B}$ denotes the Borel-$sigma$-Algebra generated from the open sets on $[0,infty)$ and {$infty$}.
I think I am going in the correct direction of one implication:
Let $f$ be measurable $rightarrow$ $f$ is a nonnegative measurable function, we have shown there exist simple function $f_n$ with $f_n rightarrow f $ for $n rightarrow infty$ where $f_n =sum_{k=1}^{n2^n} (k-1)2^{-n} mathcal{1}_{(k-1)2^{-n}le f < k2^{-n}} + nmathcal{1}_{n le f}$ with $mathcal{1}$ being the indicator-function.
Define
$A_n$ :={$(omega,y) inOmegatimesmathbb{R} | 0 < y < f_n(x)$}= $bigcup_{k=1}^{n*2^n}$ {$fin[(k-1)2^{-n},k2^{-n}),yin[0,(k-1)2^{-n})$}. $A$ is the union of those $A_n$. Is it correct, that the set where f is in is the inverse image of an interval under a measurable function, hence measurable and the set in which y is an element of is in $mathcal{B}$, which would imply $A_n$ in $sigma(mathcal{F} times mathcal{B})$, so ultimately $A$ because it is the countable union of those $A_n$.
For the other implication I could not come up with anything good so far. Because I do not really know what $mathcal{F}$ is, I will have to use
$f$ is $mathcal{F} - mathcal{B}$ measurable if $f^{-1}(mathcal{B})subset mathcal{F}$.
Maybe I am overlooking something, but how would one show this ?
measure-theory measurable-functions
measure-theory measurable-functions
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
asked Jan 9 at 0:32
babemcnuggetsbabemcnuggets
507
507
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $A$ is measurable (meaning it belongs to the product sigma algebra) then $Acap (Omega times (t,infty))$ is measurable . Its section ${omega:(omega,y) in Omega times (t,infty) text {for some}, y}$ is measurable. Hence $f^{-1}(t, infty)$ is measurable for each $t$ and $f$ is therefore measurable.
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
$endgroup$
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
add a comment |
$begingroup$
$Rightarrow:$ Suppose that $f$ is $mathcal{F}/mathcal{B}$-measurable.
We assert that
$$
A=bigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r).
$$
For, let $(omega,y)in A$, then $0<y<f(omega)$. By density of
$mathbb{Q}$, there exists $r_{0}inmathbb{Q}$ such that $y<r_{0}<f(omega)$.
Then $omegain f^{-1}left((r_{0},infty]right)$ and $yin(0,r_{0})$.
That is, $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})subseteqbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$.
Conversely, let $(omega,y)inbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$,
then there exists $r_{0}inmathbb{Q}$ such that $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})$.
Therefore $f(omega)>r_{0}$ and $0<y<r_{0}$ and hence $(omega,y)in A$.
For each $rinmathbb{Q}$, $f^{-1}left((r,infty]right)times(0,r)inmathcal{F}otimesmathcal{B}$,
where $mathcal{F}otimesmathcal{B}$ is the product $sigma$-algebra.
Now it is clear that $Ainmathcal{F}otimesmathcal{B}$ because
the union is a countable union.
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
add a comment |
$begingroup$
For the converse and Murphy's proof, let me elaborate it a little
bit, especially about the measurability of sections.
Let $(X_{1},mathcal{F}_{1})$ and $(X_{2},mathcal{F}_{2})$ be measurable
spaces. Let $X=X_{1}times X_{2}$. For $i=1,2$, let $pi_{i}:Xrightarrow X_{i}$
be the canonical projection map defined by $pi_{i}(x_{1},x_{2})=x_{i}.$
Let $mathcal{F}=mathcal{F}_{1}otimesmathcal{F}_{2}$ be the product
$sigma$-algebra. It is fundamental that $mathcal{F}$ is the smallest
$sigma$-algebra on $X$ such that for each $i$, $pi_{i}$ is $mathcal{F}/mathcal{F}_{i}$-measurable.
Moreover, if $(Y,mathcal{G})$ is a measurable space and $f:Yrightarrow X$
is a map. Then $f$ is $mathcal{G}/mathcal{F}$-measureble iff for
each $i$, the composited map $pi_{i}circ f:Yrightarrow X_{i}$
is $mathcal{G}/mathcal{F}_{i}$-measurable. This is known as the
universal property of product $sigma$-algebra.
Now let $a_{1}in X_{1}$ and $a_{2}in X_{2}$. Define $iota_{a_{1}}:X_{2}rightarrow X$
and $iota_{a_{2}}:X_{1}rightarrow X$ by $iota_{a_{2}}(x_{1})=(x_{1},a_{2})$
and $iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $iota_{a_{1}}$
is $mathcal{F}_{2}/mathcal{F}$-measurable and $iota_{a_{2}}$
is $mathcal{F}_{1}/mathcal{F}$-measurable. For example, note that
$pi_{1}circiota_{a_{2}}(x_{1})=x_{1}$, i.e., $pi_{1}circiota_{a_{2}}=id_{X_{1}}$,
the identity map on $X_{1}$, which is clearly $mathcal{F}_{1}/mathcal{F}_{1}$-measurable.
$pi_{2}circiota_{a_{2}}:X_{1}rightarrow X_{2}$ is the constant
map $x_{1}mapsto a_{2}$, which is also $mathcal{F}_{1}/mathcal{F}_{2}$-measurable.
By the universial property of product $sigma$-algebra, it follows
that $iota_{a_{2}}:X_{1}rightarrow X$ is $mathcal{F}_{1}/mathcal{F}$-measurable.
Similarly, $iota_{a_{1}}:X_{2}rightarrow X$ is $mathcal{F}_{2}/mathcal{F}$-measurable.
If $Ainmathcal{F},$ then $iota_{a_{1}}^{-1}(A)inmathcal{F}_{2}$
and $iota_{a_{2}}^{-1}(A)inmathcal{F}_{1}$. However, by writing
out $iota_{a_{1}}^{-1}(A)$, we have $iota_{a_{1}}^{-1}(A)={x_{2}mid(a_{1},x_2)in A}$.
Similarly, $iota_{a_{2}}^{-1}(A)={x_{1}mid(x_{1},a_{2})in A}$.
We are now ready to prove the converse $Leftarrow:$ Suppose that
$Ainmathcal{F}otimesmathcal{B}$. Let $t>0$ be arbitrary. Define
$iota_{t}:OmegarightarrowOmegatimesmathbb{R}$ by $iota_{t}(omega)=(omega,t)$.
By the above discussion, $iota_{t}^{-1}(A)inmathcal{F}$. But
begin{eqnarray*}
iota_{t}^{-1}(A) & = & {omegamid(omega,t)in A}\
& = & {omegamid0<t<f(omega)}\
& = & f^{-1}left((t,infty]right).
end{eqnarray*}
Moreover, $f^{-1}left((0,infty]right)=bigcup_{n=1}^{infty}f^{-1}left((frac{1}{n},infty]right)inmathcal{F}$.
Finally, if $t<0$, we have $f^{-1}left((t,infty]right)=Omegainmathcal{B}$
because $f$ is non-negative.
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
If $A$ is measurable (meaning it belongs to the product sigma algebra) then $Acap (Omega times (t,infty))$ is measurable . Its section ${omega:(omega,y) in Omega times (t,infty) text {for some}, y}$ is measurable. Hence $f^{-1}(t, infty)$ is measurable for each $t$ and $f$ is therefore measurable.
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
$endgroup$
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
add a comment |
$begingroup$
If $A$ is measurable (meaning it belongs to the product sigma algebra) then $Acap (Omega times (t,infty))$ is measurable . Its section ${omega:(omega,y) in Omega times (t,infty) text {for some}, y}$ is measurable. Hence $f^{-1}(t, infty)$ is measurable for each $t$ and $f$ is therefore measurable.
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
$endgroup$
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
add a comment |
$begingroup$
If $A$ is measurable (meaning it belongs to the product sigma algebra) then $Acap (Omega times (t,infty))$ is measurable . Its section ${omega:(omega,y) in Omega times (t,infty) text {for some}, y}$ is measurable. Hence $f^{-1}(t, infty)$ is measurable for each $t$ and $f$ is therefore measurable.
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
$endgroup$
If $A$ is measurable (meaning it belongs to the product sigma algebra) then $Acap (Omega times (t,infty))$ is measurable . Its section ${omega:(omega,y) in Omega times (t,infty) text {for some}, y}$ is measurable. Hence $f^{-1}(t, infty)$ is measurable for each $t$ and $f$ is therefore measurable.
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
answered Jan 9 at 0:46
Kavi Rama MurthyKavi Rama Murthy
57.4k42160
57.4k42160
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
add a comment |
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
$begingroup$
After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short.
$endgroup$
– babemcnuggets
Jan 9 at 1:47
add a comment |
$begingroup$
$Rightarrow:$ Suppose that $f$ is $mathcal{F}/mathcal{B}$-measurable.
We assert that
$$
A=bigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r).
$$
For, let $(omega,y)in A$, then $0<y<f(omega)$. By density of
$mathbb{Q}$, there exists $r_{0}inmathbb{Q}$ such that $y<r_{0}<f(omega)$.
Then $omegain f^{-1}left((r_{0},infty]right)$ and $yin(0,r_{0})$.
That is, $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})subseteqbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$.
Conversely, let $(omega,y)inbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$,
then there exists $r_{0}inmathbb{Q}$ such that $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})$.
Therefore $f(omega)>r_{0}$ and $0<y<r_{0}$ and hence $(omega,y)in A$.
For each $rinmathbb{Q}$, $f^{-1}left((r,infty]right)times(0,r)inmathcal{F}otimesmathcal{B}$,
where $mathcal{F}otimesmathcal{B}$ is the product $sigma$-algebra.
Now it is clear that $Ainmathcal{F}otimesmathcal{B}$ because
the union is a countable union.
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
add a comment |
$begingroup$
$Rightarrow:$ Suppose that $f$ is $mathcal{F}/mathcal{B}$-measurable.
We assert that
$$
A=bigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r).
$$
For, let $(omega,y)in A$, then $0<y<f(omega)$. By density of
$mathbb{Q}$, there exists $r_{0}inmathbb{Q}$ such that $y<r_{0}<f(omega)$.
Then $omegain f^{-1}left((r_{0},infty]right)$ and $yin(0,r_{0})$.
That is, $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})subseteqbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$.
Conversely, let $(omega,y)inbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$,
then there exists $r_{0}inmathbb{Q}$ such that $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})$.
Therefore $f(omega)>r_{0}$ and $0<y<r_{0}$ and hence $(omega,y)in A$.
For each $rinmathbb{Q}$, $f^{-1}left((r,infty]right)times(0,r)inmathcal{F}otimesmathcal{B}$,
where $mathcal{F}otimesmathcal{B}$ is the product $sigma$-algebra.
Now it is clear that $Ainmathcal{F}otimesmathcal{B}$ because
the union is a countable union.
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
add a comment |
$begingroup$
$Rightarrow:$ Suppose that $f$ is $mathcal{F}/mathcal{B}$-measurable.
We assert that
$$
A=bigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r).
$$
For, let $(omega,y)in A$, then $0<y<f(omega)$. By density of
$mathbb{Q}$, there exists $r_{0}inmathbb{Q}$ such that $y<r_{0}<f(omega)$.
Then $omegain f^{-1}left((r_{0},infty]right)$ and $yin(0,r_{0})$.
That is, $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})subseteqbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$.
Conversely, let $(omega,y)inbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$,
then there exists $r_{0}inmathbb{Q}$ such that $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})$.
Therefore $f(omega)>r_{0}$ and $0<y<r_{0}$ and hence $(omega,y)in A$.
For each $rinmathbb{Q}$, $f^{-1}left((r,infty]right)times(0,r)inmathcal{F}otimesmathcal{B}$,
where $mathcal{F}otimesmathcal{B}$ is the product $sigma$-algebra.
Now it is clear that $Ainmathcal{F}otimesmathcal{B}$ because
the union is a countable union.
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
$Rightarrow:$ Suppose that $f$ is $mathcal{F}/mathcal{B}$-measurable.
We assert that
$$
A=bigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r).
$$
For, let $(omega,y)in A$, then $0<y<f(omega)$. By density of
$mathbb{Q}$, there exists $r_{0}inmathbb{Q}$ such that $y<r_{0}<f(omega)$.
Then $omegain f^{-1}left((r_{0},infty]right)$ and $yin(0,r_{0})$.
That is, $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})subseteqbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$.
Conversely, let $(omega,y)inbigcup_{rinmathbb{Q}}f^{-1}left((r,infty]right)times(0,r)$,
then there exists $r_{0}inmathbb{Q}$ such that $(omega,y)in f^{-1}left((r_{0},infty]right)times(0,r_{0})$.
Therefore $f(omega)>r_{0}$ and $0<y<r_{0}$ and hence $(omega,y)in A$.
For each $rinmathbb{Q}$, $f^{-1}left((r,infty]right)times(0,r)inmathcal{F}otimesmathcal{B}$,
where $mathcal{F}otimesmathcal{B}$ is the product $sigma$-algebra.
Now it is clear that $Ainmathcal{F}otimesmathcal{B}$ because
the union is a countable union.
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 0:56
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
2,34138
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
add a comment |
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong?
$endgroup$
– babemcnuggets
Jan 9 at 1:10
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
$begingroup$
@babemcnuggets It is hard to directly prove that $A=cup_n A_n$. (In fact, I don't know if we really have $A=cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $Ainmathcal{F}otimesmathcal{B}$ is measurable. I include a detailed proof below.
$endgroup$
– Danny Pak-Keung Chan
Jan 9 at 2:20
add a comment |
$begingroup$
For the converse and Murphy's proof, let me elaborate it a little
bit, especially about the measurability of sections.
Let $(X_{1},mathcal{F}_{1})$ and $(X_{2},mathcal{F}_{2})$ be measurable
spaces. Let $X=X_{1}times X_{2}$. For $i=1,2$, let $pi_{i}:Xrightarrow X_{i}$
be the canonical projection map defined by $pi_{i}(x_{1},x_{2})=x_{i}.$
Let $mathcal{F}=mathcal{F}_{1}otimesmathcal{F}_{2}$ be the product
$sigma$-algebra. It is fundamental that $mathcal{F}$ is the smallest
$sigma$-algebra on $X$ such that for each $i$, $pi_{i}$ is $mathcal{F}/mathcal{F}_{i}$-measurable.
Moreover, if $(Y,mathcal{G})$ is a measurable space and $f:Yrightarrow X$
is a map. Then $f$ is $mathcal{G}/mathcal{F}$-measureble iff for
each $i$, the composited map $pi_{i}circ f:Yrightarrow X_{i}$
is $mathcal{G}/mathcal{F}_{i}$-measurable. This is known as the
universal property of product $sigma$-algebra.
Now let $a_{1}in X_{1}$ and $a_{2}in X_{2}$. Define $iota_{a_{1}}:X_{2}rightarrow X$
and $iota_{a_{2}}:X_{1}rightarrow X$ by $iota_{a_{2}}(x_{1})=(x_{1},a_{2})$
and $iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $iota_{a_{1}}$
is $mathcal{F}_{2}/mathcal{F}$-measurable and $iota_{a_{2}}$
is $mathcal{F}_{1}/mathcal{F}$-measurable. For example, note that
$pi_{1}circiota_{a_{2}}(x_{1})=x_{1}$, i.e., $pi_{1}circiota_{a_{2}}=id_{X_{1}}$,
the identity map on $X_{1}$, which is clearly $mathcal{F}_{1}/mathcal{F}_{1}$-measurable.
$pi_{2}circiota_{a_{2}}:X_{1}rightarrow X_{2}$ is the constant
map $x_{1}mapsto a_{2}$, which is also $mathcal{F}_{1}/mathcal{F}_{2}$-measurable.
By the universial property of product $sigma$-algebra, it follows
that $iota_{a_{2}}:X_{1}rightarrow X$ is $mathcal{F}_{1}/mathcal{F}$-measurable.
Similarly, $iota_{a_{1}}:X_{2}rightarrow X$ is $mathcal{F}_{2}/mathcal{F}$-measurable.
If $Ainmathcal{F},$ then $iota_{a_{1}}^{-1}(A)inmathcal{F}_{2}$
and $iota_{a_{2}}^{-1}(A)inmathcal{F}_{1}$. However, by writing
out $iota_{a_{1}}^{-1}(A)$, we have $iota_{a_{1}}^{-1}(A)={x_{2}mid(a_{1},x_2)in A}$.
Similarly, $iota_{a_{2}}^{-1}(A)={x_{1}mid(x_{1},a_{2})in A}$.
We are now ready to prove the converse $Leftarrow:$ Suppose that
$Ainmathcal{F}otimesmathcal{B}$. Let $t>0$ be arbitrary. Define
$iota_{t}:OmegarightarrowOmegatimesmathbb{R}$ by $iota_{t}(omega)=(omega,t)$.
By the above discussion, $iota_{t}^{-1}(A)inmathcal{F}$. But
begin{eqnarray*}
iota_{t}^{-1}(A) & = & {omegamid(omega,t)in A}\
& = & {omegamid0<t<f(omega)}\
& = & f^{-1}left((t,infty]right).
end{eqnarray*}
Moreover, $f^{-1}left((0,infty]right)=bigcup_{n=1}^{infty}f^{-1}left((frac{1}{n},infty]right)inmathcal{F}$.
Finally, if $t<0$, we have $f^{-1}left((t,infty]right)=Omegainmathcal{B}$
because $f$ is non-negative.
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
$begingroup$
For the converse and Murphy's proof, let me elaborate it a little
bit, especially about the measurability of sections.
Let $(X_{1},mathcal{F}_{1})$ and $(X_{2},mathcal{F}_{2})$ be measurable
spaces. Let $X=X_{1}times X_{2}$. For $i=1,2$, let $pi_{i}:Xrightarrow X_{i}$
be the canonical projection map defined by $pi_{i}(x_{1},x_{2})=x_{i}.$
Let $mathcal{F}=mathcal{F}_{1}otimesmathcal{F}_{2}$ be the product
$sigma$-algebra. It is fundamental that $mathcal{F}$ is the smallest
$sigma$-algebra on $X$ such that for each $i$, $pi_{i}$ is $mathcal{F}/mathcal{F}_{i}$-measurable.
Moreover, if $(Y,mathcal{G})$ is a measurable space and $f:Yrightarrow X$
is a map. Then $f$ is $mathcal{G}/mathcal{F}$-measureble iff for
each $i$, the composited map $pi_{i}circ f:Yrightarrow X_{i}$
is $mathcal{G}/mathcal{F}_{i}$-measurable. This is known as the
universal property of product $sigma$-algebra.
Now let $a_{1}in X_{1}$ and $a_{2}in X_{2}$. Define $iota_{a_{1}}:X_{2}rightarrow X$
and $iota_{a_{2}}:X_{1}rightarrow X$ by $iota_{a_{2}}(x_{1})=(x_{1},a_{2})$
and $iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $iota_{a_{1}}$
is $mathcal{F}_{2}/mathcal{F}$-measurable and $iota_{a_{2}}$
is $mathcal{F}_{1}/mathcal{F}$-measurable. For example, note that
$pi_{1}circiota_{a_{2}}(x_{1})=x_{1}$, i.e., $pi_{1}circiota_{a_{2}}=id_{X_{1}}$,
the identity map on $X_{1}$, which is clearly $mathcal{F}_{1}/mathcal{F}_{1}$-measurable.
$pi_{2}circiota_{a_{2}}:X_{1}rightarrow X_{2}$ is the constant
map $x_{1}mapsto a_{2}$, which is also $mathcal{F}_{1}/mathcal{F}_{2}$-measurable.
By the universial property of product $sigma$-algebra, it follows
that $iota_{a_{2}}:X_{1}rightarrow X$ is $mathcal{F}_{1}/mathcal{F}$-measurable.
Similarly, $iota_{a_{1}}:X_{2}rightarrow X$ is $mathcal{F}_{2}/mathcal{F}$-measurable.
If $Ainmathcal{F},$ then $iota_{a_{1}}^{-1}(A)inmathcal{F}_{2}$
and $iota_{a_{2}}^{-1}(A)inmathcal{F}_{1}$. However, by writing
out $iota_{a_{1}}^{-1}(A)$, we have $iota_{a_{1}}^{-1}(A)={x_{2}mid(a_{1},x_2)in A}$.
Similarly, $iota_{a_{2}}^{-1}(A)={x_{1}mid(x_{1},a_{2})in A}$.
We are now ready to prove the converse $Leftarrow:$ Suppose that
$Ainmathcal{F}otimesmathcal{B}$. Let $t>0$ be arbitrary. Define
$iota_{t}:OmegarightarrowOmegatimesmathbb{R}$ by $iota_{t}(omega)=(omega,t)$.
By the above discussion, $iota_{t}^{-1}(A)inmathcal{F}$. But
begin{eqnarray*}
iota_{t}^{-1}(A) & = & {omegamid(omega,t)in A}\
& = & {omegamid0<t<f(omega)}\
& = & f^{-1}left((t,infty]right).
end{eqnarray*}
Moreover, $f^{-1}left((0,infty]right)=bigcup_{n=1}^{infty}f^{-1}left((frac{1}{n},infty]right)inmathcal{F}$.
Finally, if $t<0$, we have $f^{-1}left((t,infty]right)=Omegainmathcal{B}$
because $f$ is non-negative.
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
$begingroup$
For the converse and Murphy's proof, let me elaborate it a little
bit, especially about the measurability of sections.
Let $(X_{1},mathcal{F}_{1})$ and $(X_{2},mathcal{F}_{2})$ be measurable
spaces. Let $X=X_{1}times X_{2}$. For $i=1,2$, let $pi_{i}:Xrightarrow X_{i}$
be the canonical projection map defined by $pi_{i}(x_{1},x_{2})=x_{i}.$
Let $mathcal{F}=mathcal{F}_{1}otimesmathcal{F}_{2}$ be the product
$sigma$-algebra. It is fundamental that $mathcal{F}$ is the smallest
$sigma$-algebra on $X$ such that for each $i$, $pi_{i}$ is $mathcal{F}/mathcal{F}_{i}$-measurable.
Moreover, if $(Y,mathcal{G})$ is a measurable space and $f:Yrightarrow X$
is a map. Then $f$ is $mathcal{G}/mathcal{F}$-measureble iff for
each $i$, the composited map $pi_{i}circ f:Yrightarrow X_{i}$
is $mathcal{G}/mathcal{F}_{i}$-measurable. This is known as the
universal property of product $sigma$-algebra.
Now let $a_{1}in X_{1}$ and $a_{2}in X_{2}$. Define $iota_{a_{1}}:X_{2}rightarrow X$
and $iota_{a_{2}}:X_{1}rightarrow X$ by $iota_{a_{2}}(x_{1})=(x_{1},a_{2})$
and $iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $iota_{a_{1}}$
is $mathcal{F}_{2}/mathcal{F}$-measurable and $iota_{a_{2}}$
is $mathcal{F}_{1}/mathcal{F}$-measurable. For example, note that
$pi_{1}circiota_{a_{2}}(x_{1})=x_{1}$, i.e., $pi_{1}circiota_{a_{2}}=id_{X_{1}}$,
the identity map on $X_{1}$, which is clearly $mathcal{F}_{1}/mathcal{F}_{1}$-measurable.
$pi_{2}circiota_{a_{2}}:X_{1}rightarrow X_{2}$ is the constant
map $x_{1}mapsto a_{2}$, which is also $mathcal{F}_{1}/mathcal{F}_{2}$-measurable.
By the universial property of product $sigma$-algebra, it follows
that $iota_{a_{2}}:X_{1}rightarrow X$ is $mathcal{F}_{1}/mathcal{F}$-measurable.
Similarly, $iota_{a_{1}}:X_{2}rightarrow X$ is $mathcal{F}_{2}/mathcal{F}$-measurable.
If $Ainmathcal{F},$ then $iota_{a_{1}}^{-1}(A)inmathcal{F}_{2}$
and $iota_{a_{2}}^{-1}(A)inmathcal{F}_{1}$. However, by writing
out $iota_{a_{1}}^{-1}(A)$, we have $iota_{a_{1}}^{-1}(A)={x_{2}mid(a_{1},x_2)in A}$.
Similarly, $iota_{a_{2}}^{-1}(A)={x_{1}mid(x_{1},a_{2})in A}$.
We are now ready to prove the converse $Leftarrow:$ Suppose that
$Ainmathcal{F}otimesmathcal{B}$. Let $t>0$ be arbitrary. Define
$iota_{t}:OmegarightarrowOmegatimesmathbb{R}$ by $iota_{t}(omega)=(omega,t)$.
By the above discussion, $iota_{t}^{-1}(A)inmathcal{F}$. But
begin{eqnarray*}
iota_{t}^{-1}(A) & = & {omegamid(omega,t)in A}\
& = & {omegamid0<t<f(omega)}\
& = & f^{-1}left((t,infty]right).
end{eqnarray*}
Moreover, $f^{-1}left((0,infty]right)=bigcup_{n=1}^{infty}f^{-1}left((frac{1}{n},infty]right)inmathcal{F}$.
Finally, if $t<0$, we have $f^{-1}left((t,infty]right)=Omegainmathcal{B}$
because $f$ is non-negative.
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
For the converse and Murphy's proof, let me elaborate it a little
bit, especially about the measurability of sections.
Let $(X_{1},mathcal{F}_{1})$ and $(X_{2},mathcal{F}_{2})$ be measurable
spaces. Let $X=X_{1}times X_{2}$. For $i=1,2$, let $pi_{i}:Xrightarrow X_{i}$
be the canonical projection map defined by $pi_{i}(x_{1},x_{2})=x_{i}.$
Let $mathcal{F}=mathcal{F}_{1}otimesmathcal{F}_{2}$ be the product
$sigma$-algebra. It is fundamental that $mathcal{F}$ is the smallest
$sigma$-algebra on $X$ such that for each $i$, $pi_{i}$ is $mathcal{F}/mathcal{F}_{i}$-measurable.
Moreover, if $(Y,mathcal{G})$ is a measurable space and $f:Yrightarrow X$
is a map. Then $f$ is $mathcal{G}/mathcal{F}$-measureble iff for
each $i$, the composited map $pi_{i}circ f:Yrightarrow X_{i}$
is $mathcal{G}/mathcal{F}_{i}$-measurable. This is known as the
universal property of product $sigma$-algebra.
Now let $a_{1}in X_{1}$ and $a_{2}in X_{2}$. Define $iota_{a_{1}}:X_{2}rightarrow X$
and $iota_{a_{2}}:X_{1}rightarrow X$ by $iota_{a_{2}}(x_{1})=(x_{1},a_{2})$
and $iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $iota_{a_{1}}$
is $mathcal{F}_{2}/mathcal{F}$-measurable and $iota_{a_{2}}$
is $mathcal{F}_{1}/mathcal{F}$-measurable. For example, note that
$pi_{1}circiota_{a_{2}}(x_{1})=x_{1}$, i.e., $pi_{1}circiota_{a_{2}}=id_{X_{1}}$,
the identity map on $X_{1}$, which is clearly $mathcal{F}_{1}/mathcal{F}_{1}$-measurable.
$pi_{2}circiota_{a_{2}}:X_{1}rightarrow X_{2}$ is the constant
map $x_{1}mapsto a_{2}$, which is also $mathcal{F}_{1}/mathcal{F}_{2}$-measurable.
By the universial property of product $sigma$-algebra, it follows
that $iota_{a_{2}}:X_{1}rightarrow X$ is $mathcal{F}_{1}/mathcal{F}$-measurable.
Similarly, $iota_{a_{1}}:X_{2}rightarrow X$ is $mathcal{F}_{2}/mathcal{F}$-measurable.
If $Ainmathcal{F},$ then $iota_{a_{1}}^{-1}(A)inmathcal{F}_{2}$
and $iota_{a_{2}}^{-1}(A)inmathcal{F}_{1}$. However, by writing
out $iota_{a_{1}}^{-1}(A)$, we have $iota_{a_{1}}^{-1}(A)={x_{2}mid(a_{1},x_2)in A}$.
Similarly, $iota_{a_{2}}^{-1}(A)={x_{1}mid(x_{1},a_{2})in A}$.
We are now ready to prove the converse $Leftarrow:$ Suppose that
$Ainmathcal{F}otimesmathcal{B}$. Let $t>0$ be arbitrary. Define
$iota_{t}:OmegarightarrowOmegatimesmathbb{R}$ by $iota_{t}(omega)=(omega,t)$.
By the above discussion, $iota_{t}^{-1}(A)inmathcal{F}$. But
begin{eqnarray*}
iota_{t}^{-1}(A) & = & {omegamid(omega,t)in A}\
& = & {omegamid0<t<f(omega)}\
& = & f^{-1}left((t,infty]right).
end{eqnarray*}
Moreover, $f^{-1}left((0,infty]right)=bigcup_{n=1}^{infty}f^{-1}left((frac{1}{n},infty]right)inmathcal{F}$.
Finally, if $t<0$, we have $f^{-1}left((t,infty]right)=Omegainmathcal{B}$
because $f$ is non-negative.
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Jan 9 at 2:00
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
2,34138
add a comment |
add a comment |
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