How to show $\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu$

Question

$\{f_n\}$ are nonnegative monotonic increasing functions. Show that $$ \int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu $$

Can someone give me a hint on how to show this?

I know that I can use the monotone convergence theorem, but I dont know how.

Hint: Consider $g_{n}=\sum_{k=1}^{n}f_{k}$ – Matematleta Jun 11 '15 at 12:18 — Matematleta, Jun 11 '15 at 12:18

Chappers · Accepted Answer · 2015-06-11T12:27:00.423

2

The $f_n$ are non-negative, so $g_n = \sum_{k=1}^{n} f_k$ is monotonic increasing. Now apply the MCT: $$ \int_X \sum_{k=1}^{\infty} f_k \, d\mu = \int_X \lim_{n \to \infty} \sum_{k=1}^{n} f_k \, d\mu = \int_X \lim_{n \to \infty} g_n \, d\mu = \lim_{n \to \infty} \int_X g_n \, d\mu = \lim_{n \to \infty} \sum_{k=1}^n \int_X f_k \, d\mu, $$ where the last equality uses the (finite) linearity of the integral.

edited Jun 11 '15 at 12:27

answered Jun 11 '15 at 12:17

Chappers

67,606

This should do it, thank you. But I have one question: Shouldn't it be $f_k$ instead of $f_n$ in the last term? – kaos Jun 11 '15 at 12:25
Oh, yes. Thanks... – Chappers Jun 11 '15 at 12:26
No Problem. Case closed ;) – kaos Jun 11 '15 at 12:28

How to show $\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu$

1 Answers1