# Assuming finite Radon-Nikodym to prove sigma-finite

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?

• 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... – gangrene Oct 28 '18 at 21:42
• You are right I need to think about this! – Olof Rubin Oct 28 '18 at 21:43
• 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. – gangrene Oct 28 '18 at 22:02
• 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 $x\in A_{n}\cap B_{n}$ – Olof Rubin Oct 28 '18 at 22:04
• Alright, I agree that $C_n = A_n \cap B_n$ does the job! – gangrene Oct 28 '18 at 22:07