Scheffé's Lemma

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space and $f_n$ be a sequence of $\mu$-integrable functions that converge almost everywhere to another $\mu$-integrable function $f$. Then $f_n$ converges to $f$ in $L^1$ if and only if $\int_X \size{f_n} d\mu$ converges to $\int_X \size{f} d\mu$.

Proof of First Direction
Suppose $f_n \to f$ in $L^1$. Then

and so $f_n \to f$ in $L^1$ implies the right-hand side of this inequality goes to $0$ as $n$ grows to infinity. Hence, since the left-hand side of the inequality is non-negative, it also goes to $0$ as $n$ grows to infinity.

Proof of Reverse Direction
Suppose $\int_X \size{f_n} d\mu \to \int_X \size{f} d\mu$. We wish to show that $\int_X |f - f_n| d\mu \to 0$.

First, note that for any real $a, b$, we have $\size{a} + \size{b} - \size{a - b} \geq 0$. Therefore $\size{f(x)} + \size{f_n(x)} - \size{f(x) - f_n(x)} \geq 0$ for each $x\in X$. Thus, we may employ Fatou's lemma on the expression $\int_X \liminf_n \size{f} + \size{f_n} - \size{f - f_n} d\mu$ to yield

$$ \int_X \liminf_n \size{f} + \size{f_n} - \size{f - f_n} d\mu \leq \liminf_n \int_X \size{f} + \size{f_n} - \size{f - f_n} d\mu. \tag{1} $$

Now, the integrand left-hand side of $(1)$ equals $2\size{f}$ almost everywhere since $f_n \to f$ pointwise almost everywhere, so the integral on the left-hand side of $(1)$ is $2\int_X f d\mu$. So we may rewrite $(1)$ as

and rearranging the left and right sides of this inequality, we get $\limsup_n \int_X \size{f-f_n} d\mu \leq 0$. This implies that $\int_X \size{f-f_n} d\mu \to 0$.