# 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 $L^1$ if and only if $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.

### Corollary

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 $L^1$ if and only if $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.

## Proof

### Sufficient Condition

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

Then:

 $\ds \size {\int_X \size f \rd \mu - \int_X \size {f_n} \rd \mu}$ $\le$ $\ds \int_X \size {f - f_n} d\mu$ Reverse Triangle Inequality on $L^1$

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.

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

$\Box$

### 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 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:

$\ds 2 \int_X \size f \rd \mu$

We may thus rewrite $(1)$ as:

 $\ds 2 \int_X \size f \rd \mu$ $\le$ $\ds \liminf_n \int_X \size f + \size {f_n} - \size {f - f_n} \rd \mu$ $\ds$ $=$ $\ds \liminf_n \int_X \size f \rd \mu + \liminf_n \int_X \size {f_n} \rd \mu + \liminf_n \paren{ - \int_X \size {f - f_n} \rd \mu }$ Integral Operator is Linear $\ds$ $=$ $\ds 2 \int_x \size f \rd \mu + \liminf_n \paren{ -\int_X \size {f - f_n} \rd \mu }$ by hypothesis: $\ds \int_X \size {f_n} \rd \mu \to \int_X \size f \rd \mu$ $\ds$ $=$ $\ds 2 \int_x \size f \rd \mu - \limsup_n \int_X \size {f - f_n} \rd \mu$ Properties of limsup and liminf

Rearranging the left hand side and right hand side of this inequality, we get:

$\ds \limsup_n \int_X \size {f - f_n} \rd \mu \le 0$

This implies that:

$\ds \int_X \size {f - f_n} \rd \mu \to 0$

$\blacksquare$

## Source of Name

This entry was named for Henry Scheffé.