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?

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    $\begingroup$ Side comment: I would spend a couple of words more on the sequence $\{ A_n\}$. Of course $\mu$ and $\nu$ are $\sigma-$finite, but we do not know a priori if they are with the same "decomposition" of the space... $\endgroup$ – gangrene Oct 28 '18 at 21:42
  • $\begingroup$ You are right I need to think about this! $\endgroup$ – Olof Rubin Oct 28 '18 at 21:43
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    $\begingroup$ I do not see that $ \bigcup C_n = X $ (also you should say something about indexing, because we do not even know if $C_n$ is nonempty). But what about taking $ C_n = \bigcup_{i=1}^n A_i \cap \bigcup_{i=1}^n B_i$? This should work. $\endgroup$ – gangrene Oct 28 '18 at 22:02
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    $\begingroup$ Well both sequences are increasing and since $\cup_n A_n = X = \cup_n B_n$ we can for every $x\in X$ find $n_1$ and $n_2$ such that $x\in A_{n_1}$ and $x\in B_{n_2}$ but if $n = \max\{n_1,n_2\}$ then since the sequences are increasing $x\in A_{n}\cap B_{n}$ $\endgroup$ – Olof Rubin Oct 28 '18 at 22:04
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    $\begingroup$ Alright, I agree that $C_n = A_n \cap B_n$ does the job! $\endgroup$ – gangrene Oct 28 '18 at 22:07

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