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If $\left\{f_n\right\}$ are uniformly integrable and $f_n\overset{a.e.}{\rightarrow}f$ ($f$ measurable), is $f$ integrable? Can "uniformly integrable" be weakened to "integrable"?

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2 Answers 2

Put $f_n(x) = n(n+1)I_{(1/n+1, 1/n)}$, $x\in[0,1]$. Then $f_n\to 0$ in $[0,1]$ and the sequence of the $f_n$ is $L^1$-bounded. However, $f_n\not\rightarrow 0$ in $L^1$ as $n\to\infty$, since $\|f_n\|_1 = 1$ for all $n$.

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Thanks. I can only mark one answer as correct so i'm going to mark my own, since this was the question i was principally interested in. –  Evan Aad Aug 9 '12 at 11:55
But this is an example of some integrable functions $f_n$ such that $f_n\to f$ almost everywhere and $f$ is integrable. Hence not an answer to the question asked. –  Did Aug 9 '12 at 13:08
up vote 0 down vote accepted
  1. Yes

    $\int \left|f\right|\overset{\mathrm{a}}{\leq}\liminf\int f_n\overset{\mathrm{b}}{<}\infty$

    a. Fatou's lemma

    b. Uniform integrability $\implies$ $\sup \int f_n<\infty$ (e.g. Klenke, Theorem 6.24i)

  2. No, e.g. $f_n:=\mathbb{1}_{\left[-n,n\right]}$ (borrowed from Per Manne's comment below)

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The answer to the second question is no. If $f_n(x)$ is the indicator function of the interval $[0,n]$ then each $f_n$ is integrable, but the limit is not. –  Per Manne Aug 9 '12 at 11:39
@PerManne: Thanks. –  Evan Aad Aug 9 '12 at 11:54
what is definition of uniformly integrable? –  nim Nov 29 '13 at 10:55

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