I am self-studying measure theory and I come across the following question:
Prove: Let $(X,\mathscr{A},\mu)$ be a measure space, let $f$ belong to $\mathscr{L}^1(X,\mathscr{A},\mu,\mathbb{R})$, and define a function $\nu$ on $\mathscr{A}$ by $\nu(A) = \int_Afd\mu$. Then $\nu$ is a signed measure on $(X,\mathscr{A})$.
Here is my attempt:
Since $f\in\mathscr{L}^1(X,\mathscr{A},\mu,\mathbb{R})$, $f$ is integrable. Let $\{A_i\}$ be a sequence of disjoint sets in $\mathscr{A}$. For each positive integer $n$ define function $f_n$ by $f_n=f\chi_{\bigcup_{i=1}^nA_i}$. We have that $|f|$ is a $[0,+\infty]$-valued integrable function on $X$, $f\chi_{\bigcup_{i=1}^{\infty}A_i}$ and $f_1,f_2,\dots$ are $[-\infty,+\infty]$-valued $\mathscr{A}$-measurable functions on $X$ such that$$f(x)\chi_{\bigcup_{i=1}^{\infty}A_i}(x) = \sum_{i=1}^{\infty}f(x)\chi_{A_i}(x) = \lim_{n\to\infty}\sum_{i=1}^nf(x)\chi_{A_i}(x) =\lim_{n\to\infty}f_n(x)\tag1$$and$$|f_n(x)| = \left|f(x)\chi_{\bigcup_{i=1}^nA_i}(x)\right| \leq |f(x)|\tag2$$hold for all $x$ in $X$. So by Lebesgue's Dominated Convergence Theorem, we have\begin{align*}\int f\chi_{\bigcup_{i=1}^{\infty}A_i}d\mu &= \lim_{n\to\infty}\int f\chi_{\bigcup_{i=1}^nA_i}d\mu\\&= \lim_{n\to\infty}\int\left(\sum_{i=1}^nf\chi_{A_i}\right)d\mu\\&= \lim_{n\to\infty}\sum_{i=1}^n\int f\chi_{A_i}d\mu\tag{by the linearity of integral}\\&= \lim_{n\to\infty}\sum_{i=1}^n\int_{A_i}fd\mu\\&= \lim_{n\to\infty}\sum_{i=1}^n\nu(A_i)\\&= \sum_{i=1}^{\infty}\nu(A_i).\end{align*}Moreover, clearly, $\nu(\emptyset)=0$. We are done.
I'm not sure if my proof is correct or not. Especially, the last step where I wrote $\lim_{n\to\infty}\sum_{i=1}^n\nu(A_i) = \sum_{i=1}^{\infty}\nu(A_i)$. I don't know whether it is legit to say that without proving the existence of the limit of the series $\sum_{i=1}^{\infty}\nu(A_i)$. Could you someone please help me check my work? I really appreciate it!