# Using Fatou's Lemma to show that $f$ is integrable.

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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$$\epsilon \geq \liminf_{n \rightarrow \infty} \int_E |f_n|\geq \int_E \liminf_{n \rightarrow \infty} |f_n|=\int_E |f|$$
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 $\int_E |f_n|d\mu=\varepsilon(1+n^{-1})$.
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