I am self-studying measure theory and got stuck on part of the proof of the Lebesgue's Dominated Convergence Theorem:
Theorem$\quad$ 2.4.5$\quad$ (Lebesgue's Dominated Convergence Theorem) Let $(X,\mathscr{A},\mu)$ be a measure space, let $g$ be a $[0,+\infty]$-valued integrable function on $X$, and let $f$ and $f_1,f_2,\dots$ be $[-\infty,+\infty]$-valued $\mathscr{A}$-measurable functions on $X$ such that \begin{align} f(x) = \lim_{n\to\infty}f_n(x)\tag1 \end{align} and \begin{align} |f_n(x)| \leq g(x),\ n=1,2,\dots\tag2 \end{align} hold at $\mu$-almost every $x$ in $X$. Then $f$ and $f_1,f_2,\dots$ are integrable, and $\int fd\mu = \lim_{n\to\infty}\int f_nd\mu$.
When proving the theorem, the book claims that "the integrability of $f$ and $f_1,f_2,\dots$ follows from that of $g$; see Proposition 2.3.8, Proposition 2.3.9, and part (c) of Proposition 2.3.4." (I added these results at the bottom of this post.)
I could not see how the integrability of $f$ and $f_1,f_2,\dots$ follow immediately from that of $g$ by applying these results. The most important reason is that the relationship (1) and (2) does not hold at every $x$ in $X$, but instead, hold at $\mu$-almost every $x$ in $X$, which makes it illegal to directly apply Proposition 2.3.4 (c). So, I tried to prove it myself.
Here are my questions for this post:
1. Is my proof of the integrability of $\mathbf{f}$ and $\mathbf{f_1,f_2,\dots}$ correct? (For example, I have doubts about the relation $\left\{x\in X:\lim_{n\to\infty}|f_n(x)| > g(x)\right\} \subseteq \bigcup_{n=1}^{\infty}\left\{x\in X:|f_n(x)|>g(x)\right\}$ I used in my proof.)
2. Am I overcomplicating anything here? (I am worried about this because the book's claim sounds like the proof shouldn't be that complicated.)
3. From my attempt below and the rest of proof of $\mathbf{\int fd\mu = \lim_{n\to\infty}\int f_nd\mu}$ presented in the book, it seems that we are not allow to conclude that $\mathbf{\int\lim_{n\to\infty}f_nd\mu = \lim_{n\to\infty}\int f_nd\mu}$. But why?
Here is my attempt to prove the integrability of $f$ and $f_1,f_2,\dots$:
Proof of the integrability of $\mathbf{f}$ and $\mathbf{f_1,f_2,\dots}$$\quad$
We first prove the integrability of $\mathbf{f_1,f_2,\dots}$. Let $n$ be a positive integer. The fact that $|f_n(x)|\leq g(x)$ holds almost everywhere implies that there is an $N\in\mathscr{A}$ such that $\mu(N)=0$ and $\{x\in X:|f_n(x)|>g(x)\}\subseteq N$. Since $|f_n|$ and $g$ are $[0,+\infty]$-valued $\mathscr{A}$-measurable functions, the set $\{x\in X:|f_n(x)|>g(x)\} \in \mathscr{A}$ by Proposition 2.1.3. So, we can let $N = \{x\in X:|f_n(x)|>g(x)\}$ and conclude that $\mu(\{x\in X:|f_n(x)|>g(x)\})=0$. Since $g$ is integrable and $[0,+\infty]$-valued, it follows that $\int g^+d\mu = \int gd\mu < +\infty$. Assume to the contrary that $\int|f_n|d\mu = \sup\left\{\int hd\mu:h \in \mathscr{S}_+\ \text{and}\ h\leq|f_n|\right\} = +\infty$. Then, as $\int gd\mu<+\infty$, there exists an $h\in\mathscr{S}_+$ such that $h\leq|f_n|$ and $\int hd\mu > \int gd\mu$. Denote this $h$ by $h=\sum_{i=1}^ma_i\chi_{A_i}$, where $a_1,\dots,a_m$ are nonnegative real numbers and $A_1,\dots,A_m$ are disjoint subsets of $X$ that belong to $\mathscr{A}$. Then \begin{align} \int|f_n|d\mu \geq \int hd\mu = \sum_{i=1}^ma_i\mu(A_i) > \int gd\mu = \sup\left\{\int pd\mu:p\in\mathscr{S}_+\ \text{and}\ p\leq g\right\}. \end{align} The set $\{x\in X:h(x)>g(x)\}\neq\emptyset$, for otherwise $h(x)\leq g(x)$ for all $x$ in $X$ would imply $\int hd\mu \leq \int gd\mu$ (see Proposition 2.3.4 (c)). Without loss of generality, suppose $\{x\in X:h(x) > g(x)\} = \bigcup_{i=1}^kA_i$ and $\{x\in X:h(x) \leq g(x)\} = \bigcup_{i=k+1}^mA_i$. So for any $x\in A_i$, $i=k+1,\dots,m$, we have $a_i\leq g(x)$. Define a function $p\in\mathscr{S}_+$ be such that $p(x) = a_i$ for all $x\in A_i$ where $i=k+1,\dots,m$, and $p(x)=0$ for all $x\in A_i$ where $i=1,\dots,k$. Then $p\leq g$ and $\int pd\mu = \sum_{i=1}^ma_i\mu(A_i)$. Now, if $\mu(\{x\in X:h(x)>g(x)\}) = \mu(\bigcup_{i=1}^kA_i) = \bigcup_{i=1}^k\mu(A_i) = 0$, then $\mu(A_i)=0$ for $i=1,\dots,k$, so that $\int hd\mu = \sum_{i=1}^ma_i\mu(A_i) = \int pd\mu \leq \int gd\mu$, contradicting the fact that $\int hd\mu > \int gd\mu$. Thus, $\mu(\{x\in X:h(x)>g(x)\}) > 0$. But $\{x\in X:h(x)>g(x)\}\subseteq\{x\in X:|f_n(x)|>g(x)\}$. So we get $\mu(\{x\in X:|f_n(x)|>g(x)\})>\mu(\{x\in X:h(x)>g(x)\})>0$, contradicting the fact that $\mu(\{x\in X:|f_n(x)|>g(x)\})=0$. Therefore, \begin{align} \int|f_n|d\mu = \sup\left\{\int hd\mu:h\in\mathscr{S}_+\ \text{and}\ h\leq|f_n|\right\} < +\infty. \end{align} Hence, $|f_n|$ is integrable. By Proposition 2.3.8, $f_n$ is integrable.
Next, we prove the integrability of $\mathbf{f}$. By Proposition 2.1.5, $\lim_{n\to\infty}|f_n|$ and $g$ are $[0,+\infty]$-valued $\mathscr{A}$-measurable functions, it follows that $\{x\in X:\lim_{n\to\infty}|f_n(x)|>g(x)\}\in\mathscr{A}$. Since $\{x\in X:\lim_{n\to\infty}|f_n(x)|>g(x)\}\subseteq\bigcup_{n=1}^{\infty}\{x\in X:|f_n(x)|>g(x)\}$ and $\mu(\{x\in X:|f_n(x)|>g(x)\})=0$ for all $n\in\mathbb{N}$, it follows that $\mu(\{x\in X:\lim_{n\to\infty}|f_n(x)|>g(x)\})=0$. Thus, $\lim_{n\to\infty}|f_n|\leq g$ holds $\mu$-almost everywhere. Then, by a similar argument, we can conclude that \begin{align} \int\lim_{n\to\infty}|f_n|d\mu = \sup\left\{\int d\mu:h\in\mathscr{S}_+\ \text{and}\ h \leq \lim_{n\to\infty}|f_n|\right\}<+\infty. \end{align} Hence, $\lim_{n\to\infty}|f_n| = |\lim_{n\to\infty}f_n|$ is integrable. By Proposition 2.3.8, $\lim_{n\to\infty}f_n$ is integrable. Since $f(x) = \lim_{n\to\infty}f_n(x)$ holds at $\mu$-almost every $x$ in $X$, by Proposition 2.3.9, $\int fd\mu$ exists and $\int fd\mu = \int\lim_{n\to\infty}f_n(x)d\mu$ which is finite given the integrability of $\lim_{n\to\infty}f_n(x)$. This implies $f$ is integrable.
Thank you very much for any help!
Results used in my attempt:
Proposition 2.1.3$\quad$ Let $(X,\mathscr{A})$ be a measurable space, let $A$ be a subset of $X$ that belongs to $\mathscr{A}$, and let $f$ and $g$ be $[-\infty,+\infty]$-valued measurable functions on $A$. Then the sets $\{x \in A:f(x) < g(x)\}$, $\{x \in A:f(x) \leq g(x)\}$, and $\{x \in A:f(x) = g(x)\}$ belong to $\mathscr{A}$.
Proposition 2.1.5$\quad$ Let $(X,\mathscr{A})$ be a measurable space, let $A$ be a subset of $X$ that belongs to $\mathscr{A}$, and let $\{f_n\}$ be a sequence of $[-\infty,+\infty]$-valued measurable functions on $A$. Then
(a) the functions $\sup_nf_n$ and $\inf_nf_n$ are measurable,
(b) the functions $\limsup_{n\to\infty}f_n$ and $\liminf_{n\to\infty}f_n$ are measurable, and
(c) the function $\lim_{n\to\infty}f_n$ (whose domain is $\{x\in A:\limsup_{n\to\infty}f_n(x)=\liminf_{n\to\infty}f_n(x)\}$) is measurable.
Proposition 2.3.4$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, let $f$ and $g$ be $[0,+\infty]$-valued $\mathscr{A}$-measurable functions on $X$m and let $\alpha$ be a nonnegative real number. Then
(a) $\int \alpha fd\mu = \alpha\int fd\mu$,
(b) $\int(f+g)d\mu = \int fd\mu+\int gd\mu$, and
(c) if $f(x)\leq g(x)$ holds at each $x$ in $X$, then $\int fd\mu \leq \int gd\mu$.
Proposition 2.3.8$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, and let $f$ be a $[-\infty,+\infty]$-valued $\mathscr{A}$-measurable function on $X$. Then $f$ is integrable if and only if $|f|$ is integrable. If these functions are integrable, then $|\int fd\mu|\leq\int|f|d\mu$.
Proposition 2.3.9$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, and let $f$ and $g$ be $[-\infty,+\infty]$-valued $\mathscr{A}$-measurable functions on $X$ that agree almost everywhere. If either $\int fd\mu$ or $\int gd\mu$ exists, then both exist, and $\int fd\mu = \int gd\mu$.
Reference$\quad$ Measure Theory by Donald Cohn.
