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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. – Asaf Karagila Mar 9 '12 at 19:47
good point--thanks for fixing. – ShawnD Mar 9 '12 at 19:51
@ShawnD: how about the integrability of $f$? – Jacob Mar 9 '12 at 20:07
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. – ShawnD Mar 9 '12 at 20:14
@ShawnD: same definition as yours. $f$ is also measurable. I forgot to mention. – Jacob Mar 9 '12 at 20:33

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? – Jacob Mar 9 '12 at 19:59
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. – Davide Giraudo Mar 9 '12 at 20:02
okay. So under what conditions will $f$ be integrable? – Jacob Mar 9 '12 at 20:07
If the $f_n$ are uniformly integrable, it's enough. – Davide Giraudo Mar 9 '12 at 20:11

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