This is a sufficient condition for something that resembles “Weak convergence”: for all

Suppose that a.e.\ and that , . If , we have for all , . Note that the result is false if .

**Proof:**

**(Case: , where is the domain of integration).**

We may assume , , otherwise the result is trivially true. Also, by Fatou’s Lemma,

Let . Since , so and there exists such that for any measurable subset with , .

Since a.e.\ ( is finite a.e.\ since ), by Egorov’s Theorem there exists closed such that and converge uniformly to on . That is, there exists such that for , for all .

Then for ,

Since is arbitrary, this means .

**(Case: ). Error: See correction below.**

Define , where is the ball with radius centered at the origin. Then , so there exists such that for , .

Since on , by Monotone Convergence Theorem,

Thus there exists such that for , .

Then for ,

so that .

**(Show that the result is false if ).**

Let . Then a.e., where . Note that , so that . Similarly, .

However if , for all but .

**Correction for the case :**

Define , where is the ball with radius centered at the origin.

Since on , by Monotone Convergence Theorem,

Thus there exists such that .

Since , by the finite measure case there exists such that for ,

So for ,

so that .

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