Scheffé's Lemma

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f_n$ be a sequence of $\mu$-integrable functions that converge almost everywhere to another $\mu$-integrable function $f$.

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

Sufficient Condition
Let $f_n \to f$ in $L^1$.

Then:

and so $f_n \to f$ in $L^1$ implies the of this inequality goes to $0$ as $n$ grows to infinity.

Since the of the inequality is non-negative, it also goes to $0$ as $n$ grows to infinity.

Necessary Condition
Let $\ds \int_X \size {f_n} \rd \mu \to \int_X \size f \rd \mu$.

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

First, by Triangle Inequality:
 * $\forall a, b \in \R: \size a + \size b - \size {a - b} \ge 0$

Therefore, for each $x \in X$:
 * $\size {\map f x} + \size {\map {f_n} x} - \size {\map f x - \map {f_n} x} \ge 0$

Thus, we may employ Fatou's Lemma for Integrals on the expression $\ds \int_X \liminf_n \size f + \size {f_n} - \size {f - f_n} \rd \mu$ to yield:


 * $(1): \ds \int_X \liminf_n \size f + \size {f_n} - \size {f - f_n} \rd \mu \le \liminf_n \int_X \size f + \size {f_n} - \size {f - f_n} \rd \mu$

Now, the integrand on the of $(1)$ equals $2 \size f$ almost everywhere since $f_n \to f$ pointwise almost everywhere.

So the integral on the of $(1)$ is:
 * $\ds 2 \int_X \size f \rd \mu$

We may thus rewrite $(1)$ as:

Rearranging the and  of this inequality, we get:
 * $\ds \limsup_n \int_X \size {f - f_n} \rd \mu \le 0$

This implies that:
 * $\ds \int_X \size {f - f_n} \rd \mu \to 0$