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 $\int_E |f| < \epsilon$.
Using Fatou's Lemma to show that $f$ is integrable.
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real-analysis
2 Answers
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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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0ok, 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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0If the $f_n$ are uniformly integrable, it's enough. – 2012-03-09