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Let $\{f_n\}$ a sequence of measurable non-negative functions on $\mathbb{R}$ converging point-wise on $\mathbb{R}$ to $f$, and let $f$ integrable over $\mathbb{R}$. If $\displaystyle \int_{\mathbb{R}} = \lim_{n \to \infty} \int_{\mathbb{R}}f_n$ then show that $\displaystyle \int_{E} f = \lim_{n \to \infty} \int_{E}f_n$ for any measurable set $E$.

One side of the inequality is trivial by Fatou's lemma. I am seeking to prove that $\displaystyle \int_{E}f \ge \lim_{n \to \infty} \int_{E}f_n$.

Any suggestions? It does seem that I will need to use some convergence theorem to derive this. Am I wrong?

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The sequence of functions $f_n 1_E$ converges pointwise to $f 1_E$, and is bounded by $f$ which is integrable. Use the dominated convergence theorem. – copper.hat Feb 1 '13 at 23:47
@copper.hat How do you deduce the $f_n$ are dominated by $f$ (particularly on the set $E$)? – David Mitra Feb 1 '13 at 23:55
Dominated convergence might not work, but Fatou's lemma does! Thanks for the hint @copper.hat! – user44069 Feb 1 '13 at 23:59
My apologies. I misread. – copper.hat Feb 2 '13 at 0:23
up vote 2 down vote accepted


Fatou's Lemma gives you that $$\tag{1}\liminf\limits_{n\rightarrow\infty} \int_{E} f_n\ \ge\ \int_E f.$$

You don't know that $\lim\limits_{n\rightarrow\infty} \int_{E} f_n$ exists yet.

But, by writing
$$ \int_E f =\int_{\Bbb R} f -\int_{E^C} f, $$ use Fatou's Lemma again to show that the right hand side of $(1)$ is no smaller than $\limsup\limits_{n\rightarrow\infty}\int_E f_n.$

You will then have

$$ \liminf\limits_{n\rightarrow\infty} \int_{E} f_n\ \ge\ \int_E f\ \ge\ \limsup\limits_{n\rightarrow\infty}\int_E f_n;$$

which implies your result.

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