Scheffé's Lemma

From ProofWiki
Jump to navigation Jump to search







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 $\int_X \size{f_n} d\mu$ 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

\(\ds \size {\int_X \size {f} d\mu - \int_X \size {f_n} d\mu}\) \(\le\) \(\ds \int_X \size {f - f_n} d\mu\) by 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.

$\Box$

Proof of Reverse Direction

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

We wish to show that $\int_X \size {f - f_n} d\mu \to 0$.

First, by Triangle Inequality, for any real $a, b$, we have $\size{a} + \size{b} - \size{a - b} \ge 0$.

Therefore $\size {\map f x} + \size {\map {f_n} x} - \size {\map f x - \map {f_n} x} \ge 0$ 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

$(1): \int_X \liminf_n \size {f} + \size {f_n} - \size {f - f_n} d\mu \le \liminf_n \int_X \size {f} + \size {f_n} - \size {f - f_n} d\mu.$


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 $2\int_X \size {f} d\mu$.


We may thus rewrite $(1)$ as

\(\ds 2\int_X \size f d\mu\) \(\le\) \(\ds \liminf_n \int_X \size {f} + \size {f_n} - \size{f - f_n} d\mu\)
\(\ds \) \(=\) \(\ds \liminf_n \int_X \size {f} d\mu + \liminf_n \int_X \size {f_n} + \liminf_n - \int_X \size {f - f_n} d\mu\) by linearity of integration
\(\ds \) \(=\) \(\ds 2 \int_x \size {f} d\mu + \liminf_n -\int_X \size {f - f_n} d\mu\) by our assumption that $\int_X \size {f_n} d\mu \to \int_X \size{f} d\mu$
\(\ds \) \(=\) \(\ds 2 \int_x \size {f} d\mu - \limsup_n \int_X \size {f - f_n} d\mu\) by Properties of limsup and liminf

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

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

$\blacksquare$

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


Aiming for a contradiction, suppose 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 $\int_X \size {f - g_n} d\mu \ge \epsilon$.

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 then since $\int_X \size {h_n} d\mu \to \int_X \size {f} d\mu$ also, the main theorem above implies that $\int_X \size {f - h_n} d\mu \to 0$.

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

$\blacksquare$


Source of Name

This entry was named for Henry Scheffé.