Mixing
Measure Theory: Simple Measure Invariance Proof
$begingroup$
Let M be a metric space, $f: M rightarrow M$ be a measurable transformation and $mu$ be a measure on M. Show that $f$ preserves $mu$ if and only if
$int phi dmu =int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$.
My attempt at a proof:
First prove this direction $rightarrow$.
We assume that $f mbox{ preserves } mu$
I have proved before that If $f:M rightarrow M$ is a measurable transformation and $mu$ is a measure on $M$. Then $f$ preserves $mu$ if and only if $int phi dmu = int phi circ f dmu$ for every integrable $phi: M rightarrow mathbb{R}$. (This was easy to prove).
Thus since bounded continuous functions are dense in $L^1$, (i.e. $C(M) subset L^1(M,mu))$.
Now for the other direction. Suppose that $intphi dmu=int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$. Need to show that $f$ preserves $mu$, meaning that we need to show that $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$.
Now, I'm not sure how to show how this, relation holds $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$, where f is a continuous bounded function. I know how to do it when $f$ is just a characteristic function (then simple etc..). But, I don't know how to show it is true for arbitrary f that is continuous and bounded. Do I need to use something with linear functionals or is there a way to do it without that? Thank you very much.
real-analysis analysis probability-theory measure-theory ergodic-theory
asked Jan 23 at 3:09
kembkemb
722313
$endgroup$
add a comment |
$begingroup$
Let M be a metric space, $f: M rightarrow M$ be a measurable transformation and $mu$ be a measure on M. Show that $f$ preserves $mu$ if and only if
$int phi dmu =int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$.
My attempt at a proof:
First prove this direction $rightarrow$.
We assume that $f mbox{ preserves } mu$
I have proved before that If $f:M rightarrow M$ is a measurable transformation and $mu$ is a measure on $M$. Then $f$ preserves $mu$ if and only if $int phi dmu = int phi circ f dmu$ for every integrable $phi: M rightarrow mathbb{R}$. (This was easy to prove).
Thus since bounded continuous functions are dense in $L^1$, (i.e. $C(M) subset L^1(M,mu))$.
Now for the other direction. Suppose that $intphi dmu=int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$. Need to show that $f$ preserves $mu$, meaning that we need to show that $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$.
Now, I'm not sure how to show how this, relation holds $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$, where f is a continuous bounded function. I know how to do it when $f$ is just a characteristic function (then simple etc..). But, I don't know how to show it is true for arbitrary f that is continuous and bounded. Do I need to use something with linear functionals or is there a way to do it without that? Thank you very much.
real-analysis analysis probability-theory measure-theory ergodic-theory
asked Jan 23 at 3:09
kembkemb
722313
$endgroup$
add a comment |
$begingroup$
Let M be a metric space, $f: M rightarrow M$ be a measurable transformation and $mu$ be a measure on M. Show that $f$ preserves $mu$ if and only if
$int phi dmu =int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$.
My attempt at a proof:
First prove this direction $rightarrow$.
We assume that $f mbox{ preserves } mu$
I have proved before that If $f:M rightarrow M$ is a measurable transformation and $mu$ is a measure on $M$. Then $f$ preserves $mu$ if and only if $int phi dmu = int phi circ f dmu$ for every integrable $phi: M rightarrow mathbb{R}$. (This was easy to prove).
Thus since bounded continuous functions are dense in $L^1$, (i.e. $C(M) subset L^1(M,mu))$.
Now for the other direction. Suppose that $intphi dmu=int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$. Need to show that $f$ preserves $mu$, meaning that we need to show that $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$.
Now, I'm not sure how to show how this, relation holds $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$, where f is a continuous bounded function. I know how to do it when $f$ is just a characteristic function (then simple etc..). But, I don't know how to show it is true for arbitrary f that is continuous and bounded. Do I need to use something with linear functionals or is there a way to do it without that? Thank you very much.
real-analysis analysis probability-theory measure-theory ergodic-theory
asked Jan 23 at 3:09
kembkemb
722313
$endgroup$
Let M be a metric space, $f: M rightarrow M$ be a measurable transformation and $mu$ be a measure on M. Show that $f$ preserves $mu$ if and only if
$int phi dmu =int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$.
My attempt at a proof:
First prove this direction $rightarrow$.
We assume that $f mbox{ preserves } mu$
I have proved before that If $f:M rightarrow M$ is a measurable transformation and $mu$ is a measure on $M$. Then $f$ preserves $mu$ if and only if $int phi dmu = int phi circ f dmu$ for every integrable $phi: M rightarrow mathbb{R}$. (This was easy to prove).
Thus since bounded continuous functions are dense in $L^1$, (i.e. $C(M) subset L^1(M,mu))$.
Now for the other direction. Suppose that $intphi dmu=int phi circ f dmu$ for every bounded continuous function $phi: M rightarrow mathbb{R}$. Need to show that $f$ preserves $mu$, meaning that we need to show that $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$.
Now, I'm not sure how to show how this, relation holds $mu (E) =mu(f^{-1}(E))$ for every measurable set $E subset M$, where f is a continuous bounded function. I know how to do it when $f$ is just a characteristic function (then simple etc..). But, I don't know how to show it is true for arbitrary f that is continuous and bounded. Do I need to use something with linear functionals or is there a way to do it without that? Thank you very much.
real-analysis analysis probability-theory measure-theory ergodic-theory
real-analysis analysis probability-theory measure-theory ergodic-theory
asked Jan 23 at 3:09
kembkemb
722313
asked Jan 23 at 3:09
kembkemb
722313
asked Jan 23 at 3:09
kembkemb
722313
asked Jan 23 at 3:09
kembkemb
722313
asked Jan 23 at 3:09
kembkemb
722313
722313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(mu)$ for any Borel measure $mu$ on any metric space?
You have assume that $mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $nu(E)=mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $mu=nu$ iff $int phi, dmu =int phi, dnu$ for all bounded continuous functions $phi$. Suppose this equation holds for such $phi$ and let $C$ be a closed set. For each $n$ let $A_n={xin M: d(x,C) <frac 1 n}$; there exists a continuous function $f_n:Mto[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x notin A_n$. Apply the hypothesis with $phi$ replaced by $f_n$ and let $ n to infty$. Note that $f_n to I_C$ pointwise. Use DCT to conclude that $mu (C)=nu (C)$. Thus $mu$ and $nu $ coincide on the class of all closed sets. Can you show that $ mu=nu$? [ One way is to use the $pi -lambda $ theorem].
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
|
show 5 more comments
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1 Answer
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$begingroup$
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(mu)$ for any Borel measure $mu$ on any metric space?
You have assume that $mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $nu(E)=mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $mu=nu$ iff $int phi, dmu =int phi, dnu$ for all bounded continuous functions $phi$. Suppose this equation holds for such $phi$ and let $C$ be a closed set. For each $n$ let $A_n={xin M: d(x,C) <frac 1 n}$; there exists a continuous function $f_n:Mto[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x notin A_n$. Apply the hypothesis with $phi$ replaced by $f_n$ and let $ n to infty$. Note that $f_n to I_C$ pointwise. Use DCT to conclude that $mu (C)=nu (C)$. Thus $mu$ and $nu $ coincide on the class of all closed sets. Can you show that $ mu=nu$? [ One way is to use the $pi -lambda $ theorem].
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
|
show 5 more comments
$begingroup$
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(mu)$ for any Borel measure $mu$ on any metric space?
You have assume that $mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $nu(E)=mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $mu=nu$ iff $int phi, dmu =int phi, dnu$ for all bounded continuous functions $phi$. Suppose this equation holds for such $phi$ and let $C$ be a closed set. For each $n$ let $A_n={xin M: d(x,C) <frac 1 n}$; there exists a continuous function $f_n:Mto[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x notin A_n$. Apply the hypothesis with $phi$ replaced by $f_n$ and let $ n to infty$. Note that $f_n to I_C$ pointwise. Use DCT to conclude that $mu (C)=nu (C)$. Thus $mu$ and $nu $ coincide on the class of all closed sets. Can you show that $ mu=nu$? [ One way is to use the $pi -lambda $ theorem].
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
|
show 5 more comments
$begingroup$
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(mu)$ for any Borel measure $mu$ on any metric space?
You have assume that $mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $nu(E)=mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $mu=nu$ iff $int phi, dmu =int phi, dnu$ for all bounded continuous functions $phi$. Suppose this equation holds for such $phi$ and let $C$ be a closed set. For each $n$ let $A_n={xin M: d(x,C) <frac 1 n}$; there exists a continuous function $f_n:Mto[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x notin A_n$. Apply the hypothesis with $phi$ replaced by $f_n$ and let $ n to infty$. Note that $f_n to I_C$ pointwise. Use DCT to conclude that $mu (C)=nu (C)$. Thus $mu$ and $nu $ coincide on the class of all closed sets. Can you show that $ mu=nu$? [ One way is to use the $pi -lambda $ theorem].
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(mu)$ for any Borel measure $mu$ on any metric space?
You have assume that $mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $nu(E)=mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $mu=nu$ iff $int phi, dmu =int phi, dnu$ for all bounded continuous functions $phi$. Suppose this equation holds for such $phi$ and let $C$ be a closed set. For each $n$ let $A_n={xin M: d(x,C) <frac 1 n}$; there exists a continuous function $f_n:Mto[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x notin A_n$. Apply the hypothesis with $phi$ replaced by $f_n$ and let $ n to infty$. Note that $f_n to I_C$ pointwise. Use DCT to conclude that $mu (C)=nu (C)$. Thus $mu$ and $nu $ coincide on the class of all closed sets. Can you show that $ mu=nu$? [ One way is to use the $pi -lambda $ theorem].
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
edited Jan 23 at 7:57
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
answered Jan 23 at 6:11
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
66.7k52867
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
|
show 5 more comments
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
Why the downvoting?
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:54
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I didn't downvote it, it was someone else.
$endgroup$
– kemb
Jan 23 at 8:08
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I upvoted it so it cancels out whoever downvoted it. @Kavi Rama Murthy I'm not familiar with $pi -lambda$ theorem. The book gives a hint. I will update my question with the hints. The idea, according to the book, is to approximate characteristic functions with continuous functions, using this fact: Let $mu$ be a probability measure on some metric space M. For, every integrable function $phi: M rightarrow mathbb{R}$, there exists a sequence $phi_n: M rightarrow mathbb{R}, n geq 1$ of uniformly continuous functions converging to $phi$ at $mu$-almost every point.
$endgroup$
– kemb
Jan 23 at 8:18
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
I know I didn't prove that bounded continuous functions are dense in $L^1(mu)$ for any Borel measure $mu$ on any metric space. But, this fact is true, right? Or are there cases where this doesn't hold. thanks.
$endgroup$
– kemb
Jan 23 at 8:20
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
$begingroup$
@kemb The best result that is known is that this property holds for a regular Borel measure on a locally compact Hausdorff space. My guess is that it is not true in a metric space, but I don't have a counterexample at this time.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:24
|
show 5 more comments
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Required, but never shown
Required, but never shown
Required, but never shown
