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I finally managed to proof this variant of the Dominated Convergence Theorem:

Theorem (Variant of Dominated Convergence Theorem). Let $f, f_k: X \to \overline{\mathbb R}$ be $\mu$-measurable, $g, g_k: X \to \overline{\mathbb R}$ be $\mu$-summable, $f_k \to f$ $\mu$-a.e., $g_k \to g$ $\mu$-a.e., $\vert f_k \vert \leq g_k$ for each $k \in \mathbb N$, $$ \lim_{k \to \infty} \int_X g_k \, \mathrm d\mu = \int_X g \, \mathrm d\mu \; .$$ Then $$ \lim_{k \to \infty} \int_X \vert f_k - f \vert \, \mathrm d\mu = 0 $$ holds.

Here is what I did: I first noted that $f_k, f \in L^1(X, \mu)$. Using Fatou's Lemma, I could show that $$ \limsup_{k \to \infty} \int_X f_k \, \mathrm d\mu \leq \int_X f \, \mathrm d\mu \leq \liminf \int_X f_k \, \mathrm d\mu \; , $$ so $\Vert f_k \Vert_{L^1} \to \Vert f \Vert_{L^1}$. Using this and some continuity properties of the integral, and Egorov's theorem, I could finally conclude that $f_k \to f$ in $L^1(X, \mu)$. I wanted to ask, if somebody knows a more direct way to show this result?

aexl
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2 Answers2

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As you already noted, it follows from Fatou's lemma that $\|f_k\|_{L^1} \to \|f\|_{L^1}$. Since we also know that $f_k$ converges almost everywhere to $f$, we get $\|f_k-f\|_{L^1} \to 0$ by applying the following well-known result.

Let $(f_k)_{k \in \mathbb{N}} \subseteq L^1$ and $f \in L^1$ such that $f_k \to f$ almost everywhere and $\|f_k\|_{L^1} \to \|f\|_{L^1}$. Then $\|f_k-f\|_{L^1} \to 0$.

This statement is not difficult to prove (e.g. using Fatou's lemma); see this question.

Community
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saz
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  • Yes, this is exactly what I did. I was just wondering, if there is a way to apply Fatou's Lemma to get $\Vert f_k - f \Vert_{L^1} \to 0$ directly. But thanks for you answer. – aexl Jan 25 '16 at 19:49
  • @aexl You mentioned Egorov's theorem and that's kind of overkill. Didn't sound to me as you did what I suggest in my answer... – saz Jan 25 '16 at 19:53
  • You're right, I used Egorov to prove this "well-known result". I saw in the thread you mentioned, that one can directly use Fatou's lemma to prove this "well-known result". I'm just wondering if there is a way to prove the theorem using Fatou's lemma without using this "well-known result". – aexl Jan 25 '16 at 20:04
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$$|f_k-f|\le |f_k|+|f|\le g_k+|f|$$

$$\int (g+|f|)\le\liminf \int ((g_k+|f|)-|f_k-f|)=\int (g+|f|)-\limsup\int|f_k-f| $$