Based on Williams' Probability w/ Martingales:
Let $(S, \Sigma, \mu)$ be a measure space.
Scheffe's Lemma Part (ii): Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s) \forall s \in S$ or a.e. in S. Then
$$\lim_{n \to \infty} \int_S |f_n – f| d\mu = 0 \iff \lim_{n \to \infty} \int_S |f_n| d\mu = \int_S |f| d\mu$$
In proving Scheffe's Lemma, we could use Fatou's Lemmas to show that
$$\lim_{n \to \infty} \int_S f_n^{+} d\mu = \int_S f^{+} d\mu$$
$$\lim_{n \to \infty} \int_S f_n^{-} d\mu = \int_S f^{-} d\mu$$
What I tried:
Fatou's Lemmas for $f_n^{+}$
$$\int_S \limsup f_n^{+} d\mu \ge \limsup \int_S f_n^{+} d\mu \ge \liminf \int_S f_n^{+} d\mu \ge \int_S \liminf f_n^{+} d\mu$$
And that's about it. I have no idea if
$$\lim_{n \to \infty} f_n^{+}(s) = f^{+}(s) \forall s \in S$$
or a.e. in S.
Is
$$\lim_{n \to \infty} \max(f_n, 0) = \max(\lim_{n \to \infty} f_n, 0)$$
?
I seem to recall from basic calculus that
$$\lim_{x \to \infty} f(g(x)) = f(\lim_{x \to \infty} g(x))$$
if $f$ is continuous.
Even if it was true, I'm not sure what I can use here. I don't think I can use monotone convergence theorem or dominated convergence theorem. Can I?
How else can I approach this?
Best Answer
The question has been edited. The updated question now asks to prove:
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 $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$