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, $\mu(X)=\infty$, $\nu(X)=1$.

  • $\begingroup$ Not sure how to fix the link works.. so I'm deleting it. And thank you! $\endgroup$ – Rubertos Apr 27 '16 at 12:22
  • $\begingroup$ @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! $\endgroup$ – 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 do 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

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

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