Using Fatou's Lemmas in proving Scheffe's Lemma Part (ii)

The question has been edited. The updated question now asks to prove:

Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s)$ a.e. in $S$. Then $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$ iff $\lim_{n \to \infty} \int_S |f_n| d\mu = \int_S |f| d\mu$

Proof: (=>) It is trivial, since, from Minkowski's inequality, we have
$$\left | \int_S|f_n|d\mu-\int_S|f|d\mu \right |\leqslant \int_S|\,|f_n| -|f|\,| d\mu\leqslant \int_S|f_n -f| d\mu$$

(<=) Note that $|f_n -f|\leqslant |f_n| +|f|$. So, for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative and using Fatou's Lemma, we have \begin{align} 2 \int_S|f|d\mu &=\int_S \lim\inf(|f_n| +|f| - |f_n -f|)d\mu \leqslant \lim\inf \int_S (|f_n| +|f| - |f_n -f|)d\mu = \\ &=\lim\inf \left (\int_S|f_n|d\mu +\int_S|f|d\mu - \int_S|f_n -f|d\mu \right) = \\ &= \left(\lim\inf\int_S|f_n|d\mu\right) +\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right) = \\ &=2\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right) \end{align} So we have $$2 \int_S|f|d\mu \leqslant 2\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right) $$ Since $f \in \mathscr{L}^1 (S, \Sigma, \mu)$ , we know that $\int_S|f|d\mu<+\infty$, and so we get $$\lim\sup\int_S|f_n -f|d\mu \leqslant 0$$ So we can conclude that $$\lim\int_S|f_n -f|d\mu = 0$$

Remark: There is another way to prove the (<=) part, which uses the Dominated Convergence Theorem (instead of Fatou's Lemma). However such way (for the question as currently stated) is a little bit "trickier" than the one presented above using Fatou's Lema. Here it is:

(<=) Consider $|f_n| \wedge |f|$ defined by $(| f_n | \wedge |f|)(x)=\min\{|f_n(x)|,|f(x)|\}$, for each $x \in \Omega$. Consider also $$ \sigma(f_n,f)(x) = \left \{\begin{aligned} &= -1 &\textrm{ if } f_n(x)f(x)<0 \\ &= 0 &\textrm{ if } f_n(x)f(x)=0 \\ &= 1 &\textrm{ if } f_n(x)f(x)>0 \end{aligned}\right.$$ for each $x \in \Omega$.

Since $\{f_n\}$ converges to $f $ a.e., we have that $\{|f_n| \wedge |f|\}$ converges to $|f|$ a.e., and $\{\sigma(f_n,f)\}$ converges to $\chi_{[f\neq 0]}$ a.e.. So, $\{\sigma(f_n,f)(|f_n| \wedge |f|)\}$ converges to $|f|$ a.e.. But we know that, for all $n$, $\vert \sigma(f_n,f)(|f_n| \wedge |f|) \vert = |f_n| \wedge |f| \leqslant |f| $ and $\int_{\Omega } |f| d\mu< \infty $. So we can apply Lebesgue Dominated Convergence Theorem and we have that $$\lim_{n \to \infty}\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu = \int_{\Omega } |f| d\mu$$ To conclude the proof, note that $$\vert f_n-f\vert = |f_n|+|f|-2\sigma(f_n,f)(|f_n| \wedge |f|)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } |f_n| d\mu +\int_{\Omega } |f| d\mu -2\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu $$ And so, since $\lim_{n \to \infty}\int_{\Omega } |f_n| d\mu = \int_{\Omega } |f| d\mu$ and $\int_{\Omega } |f| d\mu<+\infty$, we have $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

You're assuming that the $f_n$ are nonnegative, so $f_n^+ = f_n$ and you don't need to consider $f_n^-$.

First note that $$\left| \int_S f_n \, d\mu - \int_S f \, d\mu \right| \le \int_S |f_n - f| \, d\mu$$ so that one implication is easy and doesn't even require $f_n \to f$ a.e.

Next, since $|f_n - f| \le f_n + f$ you have $f + f_n - |f_n - f| \ge 0$ so that Fatou's Lemma is applicable. Since $f_n \to f$ a.e. you have $$ \int_S 2f \, d\mu = \int_S \liminf (f + f_n - |f_n - f|) \, d\mu \le \liminf \int_S f + f_n - |f_n - f| \, d\mu$$ where $$ \liminf \int_S f + f_n - |f_n - f| \, d\mu = \liminf \left[ \int_S f \, d\mu + \int_S f_n \, d\mu - \int_S |f_n - f| \, d\mu \right].$$ Next use a basic property of liminf: if $b_n \to b$, then $$\liminf (a + b_n - c_n) = a + b - \limsup c_n.$$ Thus \begin{align*}\int_S f_n \, d\mu \to \int_S f \,d\mu &\implies \liminf \int_S f + f_n - |f_n - f| \, d\mu = 2\int_S f \, d\mu - \limsup \int_S |f_n - f| \, d\mu \\ &\implies 2\int_S f \, d\mu \le 2\int_S f\, d\mu - \limsup \int_S |f_n - f| \, d\mu \\ &\implies \limsup \int_S |f_n - f| \, d\mu \le 0 \\ &\implies \int_S |f_n - f| \, d\mu \to 0.\end{align*}

Considering that $f$ is non-negative and, for each $n$, $f_n$ is non-negative, then it is simpler to prove Scheffe's lemma using Lebesgue Dominated Convergence Theorem.

(=>) Suppose $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$. Then since $$0 \leqslant \left |\int_S f_n - f d\mu \right| \leqslant \int_S |f_n - f| d\mu $$ we have $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$.

(<=) Suppose $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$. Since $\{f_n\}$ is a sequence of nonnegative Lebesgue integrable functions and $\{f_n\}$ converges to function $f $ a.e., we have that $f\geqslant 0$ a.e.. So, without loss of generality, we can assume that $f\geqslant 0$.

Now consider $f_n \wedge f$ defined by $(f_n \wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x \in \Omega$.

Since $\{f_n\}$ converges to $f $ a.e., we have that $\{f_n \wedge f\}$ converges to $f$ a.e.. But we know that, for all $n$, $\vert f_n \wedge f \vert = f_n \wedge f \leqslant f $ and $\int_{\Omega } f d\mu< \infty $. So we can apply Lebesgue Dominated Convergence Theorem and we have that $$\lim_{n \to \infty}\int_{\Omega } f_n \wedge f d\mu = \int_{\Omega } f d\mu$$ To conclute the proof, note that $$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } f_n d\mu +\int_{\Omega } f d\mu -2\int_{\Omega } (f_n\wedge f) d\mu $$ And so, since $\lim_{n \to \infty}\int_{\Omega } f_n d\mu = \int_{\Omega } f d\mu$ we have $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$