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 $\int_X \size{f} d\mu$.

Corollary

The above statement remains true after replacing convergence almost everywhere with convergence in measure.

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 $\int_X \size{f-f_n} d\mu \to 0$.

Proof of Corollary
Suppose $f_n$ converges to $f$ in measure instead. The proof of the first direction remains unchanged from the above. In the other direction, suppose $\int_X \size{f_n} d\mu \to \int_X \size{f} d\mu$; we wish to show again that $\int_X |f - f_n| d\mu \to 0$.

Suppose this is false. Then since the integral is non-negative, we can find some $\epsilon > 0$ and an infinite subsequence $g_n$ of $f_n$ such that $\int_X |f - g_n| d\mu \geq \epsilon$. But since $g_n$ still converges in measure to $f$, by Convergence_in_Measure_Implies_Convergence_a.e._of_Subsequence $g_n$ has a further subsequence $h_n$ that converges almost everywhere to $f$. But then since $\int_X \size{h_n} d\mu \to \int_X \size{f} d\mu$ also, the main theorem above implies that $\int_X |f - h_n| d\mu \to 0$. This is a contradiction, as $h_n$ is a subsequence of $g_n$, hence $\int_X |f - h_n| d\mu$ must remain larger than $\epsilon$.