# Proving the image measure $\mu_\psi$ to be $\sigma$-finite as part of the proof of a corollary of Radon-Nikodyn theorem

Definition of image measure

Let $$(X,\mathcal{A},\mu)$$ be a measure space and $$(Y,\mathcal{E})$$ be a measurable space. Let $$\psi: X\to Y$$ be a measurable function. Let $$\mu_\psi$$ denote the image measure:

$$\mu_\psi(B)=\mu(\psi^{-1}(B)), \forall B\in \mathcal{E}$$.

Radon-Nikodym theorem Let $$(X,\mathcal{A})$$ be a measurable space and let $$\mu$$ and $$\nu$$ be $$\sigma-$$finite measures over it. Let $$\nu$$ be absolutely continuous with respect to $$\mu$$. Then, there exists a measurable function $$f\ge 0$$, unique a.e. such that

$$\nu(A)=\int_A f d\mu, \forall A\in \mathcal {A}$$ with $$f$$ called the R-N derivative

Proposition Let $$(X,\mathcal{A},\mu)$$ and $$(Y,\mathcal{E},\lambda)$$ be measure spaces. Let $$\psi: X\to Y$$ be a measurable function and $$\mu_\psi$$ denote the image measure of $$\mu$$ with respect to $$\psi$$. Let $$\mu$$ and $$\lambda$$ be $$\sigma$$-finite and $$\mu_\psi$$ absolutely continuous with respect to $$\lambda$$ with Radon-Nikodyn derivative $$f_\psi=\frac{d\mu_\psi}{d\lambda}$$. Let $$g:Y\to \overline{ \mathbb{R} }$$ be a measurable function.

Then$$\int_Xg\circ\psi d\mu=\int_Ygd\mu_\psi=\int_Y g f_\psi d \lambda$$

In order to prove the above proposition,which follows directly from the R-N theorem, I need $$\mu_\psi$$ to be $$\sigma$$-finite. My lecturer said that follows from $$\mu$$ and $$\lambda$$ being $$\sigma$$-finite. But I am unable to prove it. Furthermore I've seen a number of posts where it says that the image measure of a $$\sigma$$-finite measure is not sigma finite. But in this case there's also $$\lambda$$ that is $$\sigma$$-finite, so maybe that helps? How do I prove $$\mu_\psi$$ is sigma finite?

My thoughts:

Since $$\lambda$$ is $$\sigma$$-finite, there exists a sequence $$\{B_n\}$$ of $$\mathcal{E}$$-measurable functions such that $$\lambda(B_n)<\infty$$ and $$\bigcup_n B_n=Y$$

Since $$\psi$$ is measurable, the sequence $$\psi^{-1}(B_n)$$ is made of $$\mathcal{A}$$-measurable sets such that $$\bigcup_n\psi^{-1}(B_n)=\psi^{-1}(\bigcup_n B_n)=\psi^{-1}(Y)=X$$.

So the $$\psi^{-1}(B_n)$$ covers X, but I don't think I can deduce from this that $$\mu(\psi^{-1}(B_n))<\infty$$ .How should I do it then?

• In your statement of the Radon-Nikodym theorem you write that $\nu(A) = \int_A f_\psi d\mu$, but shouldn't that be without mention of $\psi$, i.e., $\nu(A) = \int f d\mu$, where $f$ is the Radon-Nikodym derivative? I'm confused by what $\psi$ is doing there.
– Bart
Commented Aug 8, 2022 at 19:35
• I meant $\nu(A) = \int_A f d\mu$.
– Bart
Commented Aug 8, 2022 at 19:47
• @Bart . You are right, copy and paste typo., thank you. I fixed it Commented Aug 8, 2022 at 19:48

If I understand you correctly, then from the assumptions that $$\mu$$ and $$\lambda$$ are $$\sigma$$-finite and that $$\mu_\psi$$ is absolutely continuous with respect to $$\lambda$$ you want to derive that $$\mu_\psi$$ is also $$\sigma$$-finite.

But I believe that the following presents a counter-example.

Let $$X = Y = \mathbb{R}$$, let $$\mu$$ and $$\lambda$$ both be the Lebesgue measure, and let $$\psi:X\to Y$$ be defined by $$\begin{equation*} \psi(x) = x - [x] \end{equation*}$$ for every $$x\in\mathbb{R}$$, where $$[x]$$ is $$x$$ rounded down. So $$\psi$$ can be thought of as a periodic "saw-shaped" function with "teeth" that range from $$0$$ to $$1$$.

A set $$B$$ in $$Y$$ is disjoint from $$[0,1)$$ if and only if $$\psi^{-1}(B)$$ is empty. So we can restrict attention to subsets $$B$$ of $$[0,1)$$. Let $$B$$ be such a set. For any $$x$$ in $$B$$ and any integer $$n$$, $$\begin{equation*} \psi(n + x) = n + x - [n + x] = n + x - n = x, \end{equation*}$$ so $$\begin{equation*} \psi^{-1}(B) = \{n + x:x\in B, n\in\mathbb{Z}\} =\bigcup_{n\in\mathbb{Z}}\{n + x:x\in B\} . \end{equation*}$$ It follows that $$\begin{equation*} \mu_\psi(B) = \mu(\psi^{-1}(B)) = \sum_{n\in\mathbb{Z}}\mu(\{n + x:x\in B\}) = \sum_{n\in\mathbb{Z}}\mu(B) \end{equation*}$$ where in the last step I used that the Lebesgue measure $$\mu$$ is translation invariant.

If $$\lambda(B)$$ equals $$0$$ then so does $$\mu(B)$$, because both are the Lebesgue measure, and then $$\mu_\psi(B) = 0$$, and so $$\mu_\psi$$ is absolutely continuous with respect to $$\lambda$$.

But if $$\lambda(B)$$ does not equal $$0$$, then $$\mu_\psi(B)$$ is infinite. So $$\mu_\psi$$ cannot be $$\sigma$$-finite.

EDIT:

Actually, for $$\mu_\psi$$ to have a Radon-Nikodym derivative with respect to $$\lambda$$, which I think is what you are actually after, I don't think you need $$\mu_\psi$$ to be $$\sigma$$-finite. I will just quote the Radon-Nikodym theorem and Exercise 6 from chapter 4 in Donald Cohn's Measure Theory:

"Theorem 4.2.2 (Radon-Nikodym theorem) Let $$(X,\mathcal{A})$$ be a measurable space, and let $$\mu$$ and $$\nu$$ be $$\sigma$$-finite positive measures on $$(X,\mathcal{A})$$. If $$\nu$$ is absolutely continuous with respect to $$\mu$$, then there is an $$\mathcal{A}$$-measurable function $$g:X\to[0,\infty)$$ such that $$\nu(A)=\int_A gd\mu$$ holds for each $$A$$ in $$\mathcal{A}$$. The function $$g$$ is unique up to $$\mu$$-almost everywhere equality."

"Exercise 6. Show that the assumption that $$\nu$$ is $$\sigma$$-finite can be removed from Theorem 4.2.2 if $$g$$ is allowed to have values in $$[0,\infty]$$. (Hint: Reduce the general case to the case where $$\mu$$ is finite. For each positive integer $$n$$ choose a Hahn decomposition $$(P_n,N_n)$$ for the signed measure $$\nu - n\mu$$; then consider the measures $$A\mapsto \nu(A\cap(\cap_n P_n))$$ and $$A\mapsto \nu(A\cap(\cap_n P_n)^c)$$.)"

• Thanks, looks like my lecturer made a mistake in that proposition Commented Aug 9, 2022 at 21:52
• I think the right hypothesis would be to require $\mu_\psi$ to be sigma finite instead of $\mu$ in orden to apply R-N if g is not allowed to be infinite, as it happens in meaure-theoretic probability, which is the focus of my course Commented Aug 9, 2022 at 21:55