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?
Best Answer
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 $A\mapsto \nu(A\cap(\cap_n P_n)^c)$.)"