measure theory - Real Analysis, Folland Corollary 2.19 Integration of Nonnegative functions

Corollary 2.19 - If $\{f_n\}\subset L^+$, $f\in L^+$, and $f_n\rightarrow f$ a.e., then $\int f \leq \liminf\int f_n$.

Proof - We have that $\{f_n\}\subset L^+$, $f\in L^+$ and $f_n\rightarrow f$ a.e. Let $E$ be the set at which $f_n\nrightarrow f$. Therefore, $\mu(E) = 0$ by definition of almost everywhere. Then $f_n\rightarrow f$ on $X\setminus C$. By Monotone Convergence theorem, $$f_{X\setminus C} = \lim_{n\rightarrow \infty}\int_{X\setminus C}f_n = \lim_{k\rightarrow \infty}\int_{X\setminus C}\inf_{n\geq k}f_n$$ Observe that $\int_{E}f = \int_{X}f_n\chi_E = 0$ by proposition 2.16. Similarly, $\int_{E}f_n = 0$. Thus we have
\begin{align*}
\int_{X}f &= \int_{E}f + \int_{X\setminus E}f\\
&= \lim_{n\rightarrow \infty}\int_{E}f_n + \lim_{n\rightarrow \infty}\int_{X\setminus E}f_n\\

&= \lim_{n\rightarrow \infty}\int_{X}f_n\\
&= \lim_{k\rightarrow \infty}\int_{X} \inf_{n\geq k}f_n
\end{align*}
From Fatou's lemma, $$\int_{X}f = \lim_{k\rightarrow \infty}\int_{X}\inf_{n\geq k}f_n\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int f_n$$

I am not sure if this is correct, any suggestions is greatly appreciated.

Answer

Your proof is essentially OK up to close the end. In the end you can not use Fatou's lemma as you did.

I have rewritten your proof, makng minor adjustments and correcting the ending.

I also offer you a second proof. This one does not use Monotone Convergence, but actually uses Fatou's lemma.

Corollary 2.19 - If $\{f_n\}\subset L^+$, $f\in L^+$, and $f_n\rightarrow f$ a.e., then $\int f \leq \liminf\int f_n$.

Proof - We have that $\{f_n\}\subset L^+$, $f\in L^+$ and $f_n\rightarrow f$ a.e. Let $N$ be the set at which $f_n\nrightarrow f$. Therefore, then there is a set $E$ such that $N\subset E$ and $\mu(E) = 0$, by definition of almost everywhere. Then $f_n\rightarrow f$ on $X\setminus E$. Note that $\{\inf_{n\geq k}f_n\}_k $ is non decreasing sequence of positive functions in $L^+$, and $\inf_{n\geq k}f_n \to f$ in $X\setminus E$. By Monotone Convergence theorem,
$$\int f_{X\setminus E} = \lim_{k\rightarrow \infty}\int_{X\setminus E}\inf_{n\geq k}f_n$$
Observe that $\int_{E}f = \int_{X}f_n\chi_E = 0$ by proposition 2.16. Similarly, $\int_E\inf_{n\geq k}f_n = 0$, for all $k$. Thus we have

\begin{align*}
\int_{X}f &= \int_{E}f + \int_{X\setminus E}f=\\
& = \lim_{k\rightarrow \infty}\int_E\inf_{n\geq k}f_n +\lim_{k\rightarrow \infty}\int_{X\setminus E}\inf_{n\geq k}f_n\\
&= \lim_{k\rightarrow \infty}\int_{X} \inf_{n\geq k}f_n \tag{A}
\end{align*}
Since for all $k$ and $n\geq k$, we have $\inf_{n\geq k}f_n \leq f_n$, the we have
$$\int_{X} \inf_{n\geq k}f_n \leq \int f_n$$
and so, for all $k$,
$$\int_{X} \inf_{n\geq k}f_n \leq \inf_{n\geq k}\int f_n \tag{B}$$

So we have, from (A) and (B), $$\int_{X}f = \lim_{k\rightarrow \infty}\int_{X}\inf_{n\geq k}f_n\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int f_n$$

Remark: In the end we don't really apply Fatou's lemma, but we mimic part of the proof of Fatou's lemma.

Proof 2 - We have that $\{f_n\}\subset L^+$, $f\in L^+$ and $f_n\rightarrow f$ a.e. Let $N$ be the set at which $f_n\nrightarrow f$. Therefore, then there is a set $E$ such that $N\subset E$ and $\mu(E) = 0$, by definition of almost everywhere. Then $f_n\rightarrow f$ on $X\setminus E$. By Fatou's lemma,

$$ \int_{X\setminus E} f = \int_{X\setminus E} \lim_{n \to \infty} f_n= \int_{X\setminus E} \liminf_{n \to \infty} f_n \leq \liminf_{n \to \infty}\int_{X\setminus E} f_n \tag{1}$$

Observe that $\int_{E}f = \int_{X}f_n\chi_E = 0$ by proposition 2.16. Similarly, $\int_{E}f_n = 0$. Thus we have

$$\int_{X}f =\int_Ef+\int_{X\setminus E}f = 0+ \int_{X\setminus E}f= \int_{X\setminus E}f \tag{2}$$
and, for all $n$,
$$\int_{X}f_n =\int_E f_n+\int_{X\setminus E}f_n = 0+ \int_{X\setminus E}f_n= \int_{X\setminus E}f_n \tag{3}$$
Combining $(1)$, $(2)$ and $(3)$, we have

$$\int_{X}f = \int_{X\setminus E}f=\int_{X\setminus E}f\leq \liminf_{n \to \infty}\int_{X\setminus E} f_n = \liminf_{n \to \infty}\int_X f_n $$