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