# Is $\sigma$-finiteness unnecessary for Radon Nikodym theorem?

Let $(X,\mathfrak{M},\mu)$ be a $\sigma$-finite measure space and $\lambda:\mathfrak{M}\rightarrow [0,\infty]$ be a measure. If $\lambda\ll \mu$, then there exists a measurable $f:(X,\mathfrak{M})\rightarrow [0,\infty)$ such that $d\lambda= fd\mu$.

The above is the Radon Nikodym theorem stated in wikipedia. However, both texts Rudin and Folland prove the Radon Nikodym theorem under assumption that $\lambda$ is $\sigma$-finite. How do I prove the Wikipedia vesion of Radon Nikodym therem? Is there any reference?

The version on wikipedia is wrong (if it's exactly as you say; a link might have been appropriate). This is your chance to do a Good Thing by finding the Edit button and fixing it.

Counterexample with $\mu$ finite but $\nu$ not $\sigma$-finite: Let $X=\{0\}$. Define $\mu(X)=1$, $\nu(X)=\infty$.

That was easy. May as well mention that it's just as easy to give a counterexample with $\nu$ finite but $\mu$ not $\sigma$-finite: $X$ as above, $\mu(X)=\infty$, $\nu(X)=1$.

• Not sure how to fix the link works.. so I'm deleting it. And thank you! – Rubertos Apr 27 '16 at 12:22
• @David I posted a question related to your counterexample here. I didn't have a successful answer, so could you have a look at it? Thank you! – user39756 Oct 23 '16 at 13:33

EDIT: It seems that OP does not allow $$f$$ to take $$\infty$$, then the $$\sigma$$ finiteness is needed. If we allow function to take $$\infty$$, we does not need the $$\sigma$$ finiteness condition. I'll leave the answer below as an extension.

Let $$(X,\mathcal A)$$ is a measurable space, and $$\mu,\nu$$ are two measures on it, with $$\nu \ll \mu$$. We need $$\sigma$$ finiteness for $$\mu$$, and we need $$\mu$$ to be positive measure. However we do NOT need $$\sigma$$ finiteness for $$\nu$$, and actually $$\nu$$ could be of any measure, according to an exercise in the book of A course in abstract analysis by J.B. Conway; (Page 113, Exercise 3). I put a solution of mine to that exercise below:

Let $$(X,\mathcal A)$$ is a measurable space, and $$\mu$$ is a positive $$\sigma$$-finite measure on $$(X,\mathcal A)$$, and $$\nu$$ is any measure on $$(X,\mathcal A)$$ (could be positive, signed, or complex measure), and $$\nu \ll \mu$$. Then Radon-Nikodym still holds.

Now, I'm gonna provide a proof given that we've already proved Radon-Nikodym Theorem for $$\sigma$$-finite positive measure of $$\mu$$ and $$\sigma$$-finite signed measure $$\nu$$, where $$\nu \ll \mu$$.

Proof: Step 1, we consider the case that $$\mu$$ is $$\sigma$$-finite positive measure, and $$\nu$$ is signed measure. Firstly, let $$\lambda$$ be a positive finite measure, we will prove two conclusions (*1) and (*2).

Given $$n\ge 1$$, let $$\{P_n,N_n\}$$ be the Hahn decomposition for the signed measure of $$\nu-n\lambda$$, where $$P_n$$ are positive sets, and $$N_n$$ are negative sets. Thus we have $$(\nu-n\lambda)(N_n) \le 0 \Rightarrow \nu(N_n) \le n\lambda(N_n)$$. Let $$P=\cap_{n=1}^\infty P_n, N=\cup_{n=1}^\infty N_n$$. Consider $$\nu$$ restricted to set $$N$$, $$\nu|_N(E)=\nu(E\cap N)$$, we have $$\forall n \ge 1, \nu|_N(N_n)=\nu(N_n\cap N)=\nu(N_n) \le n\lambda(N_n) < \infty$$, and thus $$\nu|_n$$ is $$\sigma$$-finite (*1).

$$\forall A \in \mathcal A$$ and $$A \subset P$$, we have $$\forall n \ge 1, A \subset P_n$$, and thus $$\forall n \ge 1, (\nu-n\lambda)(A)\ge 0 \Rightarrow \nu(A) \ge n\lambda(A)$$. This means (*2): $$\nu(A)=\begin{cases}0, \mu(A) = 0 \\ \infty, \mu(A) > 0\end{cases}$$

Now $$\mu$$ is $$\sigma$$-finite, and thus we could find a series of increasing measurable sets $$F_i \uparrow X$$, s.t. $$\forall i \in \mathbb N, \mu(F_i) < \infty$$. Let $$\mu_i=\mu|_{N_i}$$, then $$\mu_i$$ are finite positive measures. We apply (*1) to every $$\mu_i$$, and get a series of $$\sigma$$-finite signed measure $$\nu_i=\nu|_{N_i}$$, where $$N_i=\cup_{n=1}^\infty N_{i,n}\subset F_i$$, and it's obviously that $$\nu_i \ll \mu_i$$.

We could apply Radon-Nikodym Theorem for $$\sigma$$-finite positive $$\mu$$, and $$\sigma$$-finite signed measure $$\nu$$, to find $$\mu_i$$-measurable function $$f_i$$, s.t. $$\forall A \in \mathcal A, A \subset N_i, ~\nu_i(A)=\int_Af_i~d\mu_i$$ Now notice that if $$i \le j$$, then $$F_i \subset F_j$$, and thus $$\mu_i(N_{i,n})=\mu(N_{i,n} \cap F_i)=\mu(N_{i,n} \cap F_j)=\mu_j(N_{i,n})$$. So we have: $$(\nu-n\mu_i)(N_{i,n}) \le 0 \Rightarrow (\nu-n\mu_i)(N_{i,n}) \le 0$$ This means that $$N_{i,n} \subset N_{j,n}$$, and $$P_{i,n} \supset P_{j,n}$$, and therefore $$N_i \uparrow N^*,~P_i \downarrow P^*, ~ N^* \cap P^* = \emptyset, ~N^* \cup P^*= X$$. Thus on the set of $$N_i$$, we have $$f_i = f_j, a.e.(\mu_i)$$. Now we can finally define our function: $$f(x)=\begin{cases}f_i(x), x \in N_i \\ \infty, x\in P^*\end{cases}$$

And we have $$\nu(A \cap N_i)=\nu_i(A)=\int_Af~d\mu_i=\int_{A\cap N_i}f~d\mu$$ Let $$i \to \infty$$, we have $$\nu(A\cap N^*)=\int_{A\cap N^*}f~d\mu$$

Next, let's consider set $$A \cap P^*)$$. Obviously, $$\forall i,~ A\cap P^*\subset P_i$$, and we apply (*2) to know that:

When $$\mu(A \cap P^*)=0$$, and recall definition of Lebesgue integration that if $$a_i=0, \mu(E_i) = \infty$$, we define $$a_i\mu(E_i) = 0$$, thus we have $$\nu(A\cap P^*)=0=\int_{A \cap P^*}f~d\mu$$ When $$\mu(A\cap P^*) > 0$$, we have $$\nu(A\cap P^*)=\infty=\int_{A\cap P^*}f~d\mu$$

Finally in summary, we have: $$\forall A \in \mathcal A, ~ \nu(A) = \int_{A\cap P^*}f~d\mu + \int_{A \cap N^*}f~d\mu = \int_Af~d\mu$$ Now we've approved Radon-Nikodym for $$\mu$$ to be $$\sigma$$-finite positive measure and $$\nu$$ to be any signed measure

Note here we assume the signed measure would take $$\infty$$, some other definition would let signed measure to take $$-\infty$$ - but it cannot take both $$\infty$$ and $$-\infty$$. For the case where signed measure would take $$-\infty$$, proof is similar.

Step 2, we consider the case that $$\mu$$ is $$\sigma$$-finite positive measure, while $$\nu$$ is complex measure. We write $$\nu=\nu_R + i\nu_I$$, where $$\nu_R,\nu_I$$ are both signed measures (to be precise, we know that they have to be finite as well), and we apply conclusion in Step1 to $$\nu_R$$ and $$\nu_I$$ respectively and add them together - we then get the Radon-Nikodym Theorem for $$\mu$$ is $$\sigma$$-finite positive measure, and $$\nu$$ is complex measure.

Thus, now we've proved Radon-Nikodym Theorem for the case when $$\mu$$ being a positive $$\sigma$$-finite measure on $$(X,\mathcal A)$$, and $$\nu$$ being any measure on $$(X,\mathcal A)$$ (could be positive, signed, or complex measure), with $$\nu \ll \mu$$. Q.E.D

• It seems here you are allowing the function $f$ to take the value $\infty$, but the statement in the question does not permit this. – Nate Eldredge Sep 2 '19 at 23:40
• @Nate Eldredge Ah, good point. – Yujie Zha Sep 2 '19 at 23:41

It is possible to prove Radon Nikodym theorem without assuming $\lambda$ is $\sigma$-finite.