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Theorem: (Monotone Convergence Theorem (MCT)). Let $(f_n)_{n\geq 1}$ be measurable functions such that $0\leq f_n\uparrow f.$ Then $f$ is measurable and $$\displaystyle\lim\limits_{n\to\infty}\int_S f_n d\mu = \int_S f d\mu$$

Exercise: Let $f_1,f_2,\ldots:S\to[0,\infty]$ be measurable functions. Then show that $$\displaystyle\int_S\sum\limits_{n=1}^\infty f_n d\mu = \sum\limits_{n=1}^\infty\int_S f_n d\mu$$

by appling the MCT on $\displaystyle s_n = \sum\limits_{j=1}^n f_j$ for $n\geq 1$.

What I've tried: I need to apply MCT on $\displaystyle s_n = \sum\limits_{j=1}^n f_j$. We have that $\displaystyle\lim\limits_{n\to\infty}\sum\limits_{j=1}^nf_j = \sum\limits_{j=1}^\infty f_j$. If we take $\sum\limits_{j=1}^\infty f_j = s$ then according to the MCT $\displaystyle\lim\limits_{n\to\infty}\int_S s_n d\mu = \int_S s d\mu$, or equivalently $$\displaystyle\lim\limits_{n\to\infty}\int_S \sum\limits_{j=1}^n f_j d\mu = \int_S \sum\limits_{n=1}^\infty f_j d\mu.$$ I'm not sure how to proceed from here and whether or not I'm on the right track.

Question: How do I solve this exercise?

Thanks in advance!

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  • $\begingroup$ Why did you not accepte the answer below? $\endgroup$
    – user503348
    Jan 3, 2018 at 13:14

1 Answer 1

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Observes that by finite sum we have $$\sum_{j=0}^{n}\int_S f_j=\int_S\sum_{j=0}^{n}f_j=\int_S s_n $$. with $$s_n=\sum_{j=0}^{n}f_j~~~~then~~~~0\leq s_n\uparrow s = \sum_{j=0}^{\infty}f_j.$$ then apply the MCT theorem to the sequence $s_n. $ since $s_n\le s_{n+1}$ therefore$$\sum_{j=0}^{\infty}\int_S f_jd\mu=\lim\limits_{n\to\infty}\sum_{j=0}^{n}\int_S f_jd\mu= \displaystyle\lim\limits_{n\to\infty}\int_S s_n d\mu = \int_S s d\mu =\int_S\sum_{j=0}^{\infty} f_jd\mu $$

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