Let $(g_n), g$ be (Lebesgue) integrable functions (i.e. $\int |g_n|, \int |g| < \infty$). Suppose $g_n\to g$ almost everywhere. Show that $\int |g_n – g|\to 0$ if and only if $\int |g_n| \to \int |g|$.
Suppose $D$ is the (common) domain of the functions.
For the only if statement, if $\int |g_n – g|\to 0$. Then $|\int |g_n| – \int |g|| \leq \int ||g_n|-|g|| \leq \int |g_n-g|\to 0.$ The other direction seems trickier. I tried using Egorov's theorem, but that only works if $m(D) < \infty$. And even if $m(D) < \infty$, we just have that for all $\epsilon > 0, \exists F\subseteq D$ so that $m(D\backslash F) < \epsilon$ and $g_n\to g$ uniformly on $F$. The Dominated convergence theorem doesn't seem to apply here either as if $\int |f|\leq \int |g|$, that doesn't necessarily imply that $|f|\leq |g|$ almost everywhere. So I can't seem to find a function $g\in L^1$ so that $|f_n|\leq g$ almost everywhere for each $n$.
Best Answer
Suppose $\int |g_n| \to \int |g|$. Since $|g_n - g| \le |g_n| + |g|$ for all $n$ and $g_n \to g$ a.e., Fatou's lemma yields $$\varliminf_{n\to \infty} \int (|g_n| + |g| - |g_n - g|) \ge \int 2|g|$$ The left-hand side of this inequality is $2\int |g| - \varlimsup_{n\to \infty} \int |g_n - g|$ since $\int |g_n| \to \int |g|$. Therefore $$2\int |g| - \varlimsup_{n\to \infty} \int |g_n - g| \ge 2\int |g|$$ Since $\int |g| < \infty$ we deduce $$\varlimsup_{n\to \infty} \int |g_n - g| \le 0$$ and the result follows.