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Let $f_n$ be a sequence of measurable functions on $\mathbb{R}$ converging a.e. to $f$. If $0\leq f_n\leq f$ a.e. Does it follow that $\displaystyle\int_\mathbb{R} f_n\ dm\to\displaystyle\int_\mathbb{R} f\ dm$?

I think this is false but I can't think of any counterexample. Also, if I put another condition that $f_n$ is a sequence of integrable functions, does this imply that $f$ would also be integrable? Hence, the conclusion will hold by the Dominated Convergence Theorem? Thanks for any response.

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  • $\begingroup$ monotone convergence theorem? $\endgroup$ – Sean Nemetz May 3 at 22:15
  • $\begingroup$ I was thinking MCT too but $f_n$ is not necessarily increasing. But can we force $f_n$ to be eventually increasing in this case since $f_n$ is always less than $f$ but $f_n$ needs to converge to $f$ at the same time? $\endgroup$ – John Thompson May 3 at 22:17
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    $\begingroup$ Hint: If you want to use MCT, consider $g_m=\inf_{n\ge m}f_n$ $\endgroup$ – fedja May 3 at 22:19
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    $\begingroup$ Another similar hint is using Fatou Lema! $\endgroup$ – HFKy May 3 at 22:42
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    $\begingroup$ It is a Beatifully lemma :D $\endgroup$ – HFKy May 3 at 22:50
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It is true. On one hand, since $0 \leq f_n \leq f$ for any $n$, one has $$\limsup_{n \to \infty} \int f_n dm \leq \int f dm.$$ On the other hand, by Fatou's lemma, we have $$\int f dm \leq \liminf_{n \to \infty} \int f_n dm.$$ Combining these two inequalities, we conclude $$\int f dm = \lim_{n \to \infty} \int f_n dm.$$ Note that here we allow both sides of the identity equal infinity.

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