I'm trying to fill in the details of the proof sketch given in my lecture note. This is to enforce where the assumptions are used.
Radon–Nikodym theorem: Let $\mu, \nu$ be $\sigma$-finite measures on a measurable space $(X, \Sigma)$. If $\nu \ll \mu$ then there is a unique (up to a $\mu$-null set) $\Sigma$-measurable function $f:X \to [0, \infty)$ such that $$ \nu (A) = \int_A f \mathrm d \mu \quad \forall A \in \Sigma. $$
Could you have a check on my attempt?
Proof:
- We first assume $\mu, \nu$ are both finite.
- Existence:
Let $F$ be the collection of all $\Sigma$-measurable function $f:X \to [0, \infty]$ such that $$ \int_A f \mathrm d \mu \le \nu(A) \quad \forall A \in \Sigma. $$
Then $0 \in F$ and thus $F \neq \emptyset$. Let $(f_n) \subset F$ such that $$ \lim_n \int_X f_n \mathrm d \mu = \sup_{f\in F} \int_X f \mathrm d \mu. $$
It's easy to verify that if $f_1, f_2 \in F$ then $\max \{f_1, f_2\} \in F$. As such, we can assume $(f_n)$ is non-decreasing. Let $$ g:= \lim_n f_n =\sup_n f_n. $$
Then $g$ takes values in $[0, \infty]$ and is $\Sigma$-measurable. By monotone convergence theorem, $g \in F$. We define a measure $\nu_0$ by $$ \nu_0 (A) := \nu(A) - \int_A g \mathrm d \mu \quad \forall A \in \Sigma. $$
Then $\nu_0$ is non-negative. We claim that $\nu_0 = 0$. Assume the contrary that $\nu_0 (X)>0$. Because $\mu$ is finite, there is $\varepsilon>0$ such that $$ \nu_0(X) > \varepsilon \mu(X). $$
Let $(P, N)$ be a Hahn decomposition of the signed measure $\nu_0 - \varepsilon \mu$. We claim that $\mu(P)>0$. If not, $\mu(P) = \nu(P)=0$ because $\nu \ll \mu$. Then $\nu_0 (P) \le \nu(P) = 0$ and we get a contradiction $(\nu_0 - \varepsilon \mu) (X) = (\nu_0 - \varepsilon \mu) (N) \le 0$. Let $g' := g + \varepsilon 1_P$. Then $$ \begin{align} \nu(A) - \int_A g' \mathrm d\mu &= \nu_0 (A) -\int_A \varepsilon 1_P \mathrm d\mu \\ &= \nu_0 (A) - \varepsilon \mu (A \cap P) \\ &\ge \nu_0 (A \cap P) - \varepsilon \mu (A \cap P) \\ &= (\nu_0 - \varepsilon \mu) (A) \ge 0. \end{align} $$
Also, $$ \int_A g' \mathrm d\mu - \int_A g \mathrm d\mu = \varepsilon \mu(P) >0. $$
This contradicts the maximality of $g$. As such, $$ \nu(A) = \int_A g \mathrm d \mu \quad \forall A \in \Sigma. $$
Because $\nu$ is finite, $\nu(X) < \infty$ and thus $g$ is $\mu$-integrable. This implies $\{g = \infty\}$ is a $\mu$-null set. As such, we obtain the required $f$ by re-defining $g$ as follows, i.e., $$ f(x)= \begin{cases} g(x) & \text {if } g(x)<\infty \\ 0 & \text {otherwise}. \end{cases} $$
- Uniqueness:
Let $f':X \to [0, \infty)$ be another function that satisfies the required properties. Let $A := \{f> f'\}$ and $B := \{f< f'\}$. Then $A, B \in \Sigma$. Hence $$ \nu(A) = \int_A f \mathrm d\mu = \int_A f' \mathrm d\mu \quad \text{and} \quad \nu(B) = \int_B f \mathrm d\mu = \int_B f' \mathrm d\mu. $$
It follows that $$ \int_A (f-f') \mathrm d\mu = \int_B (f'-f) \mathrm d\mu =0. $$
As such, $A, B$ are $\mu$-null sets.
- $\mu, \nu$ are $\sigma$-finite.
There is a partition $(A_n) \subset \Sigma$ of $X$ such that $\mu(A_n), \nu(A_n) < \infty$ for all $n$. Apply above result, we get $(f_n)$ with $f_n: A_n \to [0, \infty)$ being $\Sigma$-measurable such that $$ \nu(A \cap A_n) = \int_{A \cap A_n} f_n \mathrm d \mu \quad \forall A \in \Sigma. $$
We can extend $f_n$ to the whole $X$ by re-defining $f_n$ as follows, i.e., $$ f_n'(x)= \begin{cases} f_n(x) & \text {if } x\in A_n \\ 0 & \text {otherwise}. \end{cases} $$
Then $$ \nu(A \cap A_n) = \int_{A} f'_n \mathrm d \mu \quad \forall A \in \Sigma. $$
By monotone convergence theorem, we get $$ \nu(A) = \int_A \sum_n f'_n \mathrm d\mu \quad \forall A \in \Sigma. $$
Let $f : = \sum_n f'_n$. Notice that $f$ takes values in $[0, \infty)$. Because each $f_n$ is unique (up to a $\mu$-null set), so is $f$. This completes the proof.