Given convergence of integrand and integral, show convergence of integral over arbitrary measurable set

Question

All measures are Lebesgue.

$\forall n \in \mathbb{N}$, let $f_n: \mathbb{R} \rightarrow [0, \infty]$ be measurable and almost everywhere $f_n \rightarrow f$; moreover, suppose that $\int f_n dλ \rightarrow \int f dλ < +\infty$.

Show that, for any measurable set $A, \int_A f_n dλ \rightarrow \int_A f dλ$.

I really am not sure where to even start.

I do not see how it (immediately) applies. It does not seem to involve measurable sets. — user332420, Apr 18 '16 at 06:34

score 0 · Accepted Answer · answered Apr 18 '16 at 06:40

0

Assuming the $f_n$'s are integrable, we can apply the generalized dominated convergence theorem:

If $|f_n|\leq g_n$, $g_n$ are integrable, $g_n \rightarrow g \in L^1$ (integrable), $f_n \rightarrow f$, and $\int g_n \rightarrow \int g$, then $\int f_n \rightarrow \int f$.

The proof of this is very similar to the proof of the usual DCT.

Now just use $|f_n \chi_A |\leq |f_n|=f_n$ for any measurable $A$.

answered Apr 18 '16 at 06:40

Rick Sanchez

1,016

Does the finiteness of $\int f $ imply that for any n large enough, $\int f_n < +\infty$? – user332420 Apr 18 '16 at 13:13
Yes, if not then $\int f_n =\infty$ for infinitely many $n$, so the sequence could not converge to a finite number. – Rick Sanchez Apr 18 '16 at 14:29

Given convergence of integrand and integral, show convergence of integral over arbitrary measurable set

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