Suppose that we assume the version of Radon-Nikodym for finite measure that is if $X$ is some set, $\mathcal{A}$ is a $\sigma$-algebra on $X$ and $\mu$ and $\nu$ are finite measures on $X$ where $\nu$ is absolutely continuous with respect to $\mu$ then we can find a nonnegative integrable function such that
$$\nu(E) = \int_E gd{\mu}.$$
I was wondering if we could argue as follows in the case where $\mu$ and $\nu$ are $\sigma$-finite:
Let $\{A_n\}_n$ be a sequence over $\mathcal{A}$ such that $A_n\subseteq A_{n+1}$, $\mu(A_{n+1})<\infty$, $\nu(A_{n+1})<\infty$ and $\cup_n A_n = X$.
Then by applying the finite version of Radon-Nikodyn to the case where we restrict our measures to $A_n$ we obtain a sequence of functions $\{g_n\}_n$ where $g_n:A_n\rightarrow \mathbf{R}$ and for any $\mathcal{A}\ni E\subset A_n$
$$\nu(E) = \int_E g_n d\mu.$$
We extend each $g_n$ to be zero outside of $A_n$.
Claim: $g_n\leq g_{n+1}$ almost everywhere. Let $E = \{x\in A_n\mid g_n(x)>g_{n+1}(x)\}$ then
$$\nu(E) =\int_{E}g_{n}d\mu>\int_{E}g_{n+1}d\mu=\nu(E)$$ which implies that $\mu(E) = 0\Rightarrow \nu(E) =0$. Now define
$$g = \lim_n g_n$$
then by the monotone convergence theorem
$$\nu(E) = \lim_n \int_E g_nd\mu = \int_E gd\mu.$$
My question is whether this reasoning is valid?