Scheffé's Lemma
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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.