Scheffé's Lemma/Corollary

Corollary to Scheffé's Lemma
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f_n$ be a sequence of $\mu$-integrable functions that convergence in measure to another $\mu$-integrable function $f$.

Then $f_n$ converges to $f$ in $L^1$ $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.

Proof
Let $f_n$ converges to $f$ in measure instead.

The proof of the first direction remains unchanged from the above.

In the other direction, suppose:
 * $\ds \int_X \size {f_n} \rd \mu \to \int_X \size f \rd \mu$

We wish to show again that:
 * $\ds \int_X \size {f - f_n} \rd \mu \to 0$.

this is false.

Since the integral is non-negative, we can find some $\epsilon > 0$ and an infinite subsequence $g_n$ of $f_n$ such that:
 * $\ds \int_X \size {f - g_n} \rd \mu \ge \epsilon$

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 we also have:
 * $\ds \int_X \size {h_n} \rd \mu \to \int_X \size f \rd \mu$

Hence from Scheffé's Lemma:
 * $\ds \int_X \size {f - h_n} \rd \mu \to 0$

This is a contradiction, as $h_n$ is a subsequence of $g_n$, hence $\ds \int_X \size {f - h_n} \rd \mu$ must remain larger than $\epsilon$.