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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.

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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

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

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