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Suppose $f_n$, $g_n$, $f$ and $g$ are integrable, $f_n \to f$ almost everywhere, $g_n \to g$ almost everywhere, $|f_n| \le g_n$ for each $n$, and $\int g_n \to \int g$. Does it follow that $\int f_n \to \int f$?

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  • $\begingroup$ There is a `generalized' DCT that says almost exactly this. If $f_n,g_n,f,g\in L^1$ and both $f_n\to f$, $g_n\to g$ a.e., $|f_n|\le g_n$, and $\int g_n\to\int g$, then $\int f_n \to \int f$. Really it's not that hard to show using the methods of the proof of the vanilla DCT. $\endgroup$ – Nobody Jul 18 '16 at 3:22
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First observe that $|f|\leq g$ since $|f_n|\leq g_n$ for all $n$ and $f_n\to f,g_n\to g$ almost everywhere. Therefore the function $$ h_n=g+g_n-|f-f_n| $$ is non-negative, so we can apply Fatou's lemma to conclude that $$ \int \liminf_{n\to\infty}h_n\leq \liminf_{n\to\infty}\int h_n$$

Using the fact that $\int g_n\to \int g<\infty$, it follows that $$ \limsup_{n\to\infty}\int |f_n-f|=0 $$ and since $$ \Big|\int f_n-\int f\Big|\leq \int |f-f_n| $$ this shows that $\int f_n\to\int f$.

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