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Suppose $f_n \to f$ almost everywhere on $X$. Let $\epsilon > 0$ and choose $\delta > 0$ such that for all measurable sets $E\subseteq X$ such that $ \mu(E)< \delta $, we have $\int_E |f_n| < \epsilon$ for every $n$. Using Fatou's Lemma, how can prove that $f$ is integrable on any measurable set $E\subseteq X$ such that $\mu(E) < \delta$ and $\int_E |f| < \epsilon$.

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$$\epsilon \geq \liminf_{n \rightarrow \infty} \int_E |f_n|\geq \int_E \liminf_{n \rightarrow \infty} |f_n|=\int_E |f|$$

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    If you only write one line of math, better use two $$ symbols for a better display (it allows the LaTeX to know that you're not putting the math mid-line so there is no problem in longer symbols and placing limits below the actual line.2012-03-09
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    good point--thanks for fixing.2012-03-09
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    @ShawnD: how about the integrability of $f$?2012-03-09
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    what is your definition of the integrability of $f$ on $E$? Mine is that $\int_E |f| <\infty$. You haven't mentioned anything about measurability, but I'm assuming your $f_n$'s (and hence $f$) are all measurable.2012-03-09
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    @ShawnD: same definition as yours. $f$ is also measurable. I forgot to mention.2012-03-09
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    ok, so then we've show that $f$ is integrable on $E$, no?2012-03-09
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Since $f_n$ converges to $f$ almost everywhere, $|f_n|$ converges to $|f|$ almost everywhere, hence $$\int E|f|d\mu=\int_E\lim_n |f_n|d\mu=\int_E\liminf_n|f_n|d\mu\overset{\mbox{Fatou}}{\leq} \liminf_n\int_E|f_n|d\mu\leq \varepsilon,$$ since we have for all $n$, $\int_E |f_n|<\varepsilon$. (note that the inequality may no be strict, for example if $\int_E |f_n|d\mu=\varepsilon(1+n^{-1})$.

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    and this also shows that $f$ is integrable...right?2012-03-09
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    It depend, if you take the real line with Lebesgue measure and $f_n=f$ the constant function equal to $1$, we can find the corresponding $\delta$ for a fixed $\varepsilon$ but $f$ is not integrable.2012-03-09
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    okay. So under what conditions will $f$ be integrable?2012-03-09
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    If the $f_n$ are uniformly integrable, it's enough.2012-03-09