Prove $\int \bigg(\sum_{n=1}^{\infty} f_n \bigg) d \mu = \sum_{n=1}^{\infty} \bigg( \int f_n d \mu \bigg) $

Question

Suppose that $\{ f_n \}_{n=1}^{\infty}$ is a sequence in $L^{+}(X, \mathcal{M}).$ Prove that $$\int \bigg(\sum_{n=1}^{\infty} f_n \bigg) d \mu = \sum_{n=1}^{\infty} \bigg( \int f_n d \mu \bigg) $$

I can't seem to find anything in my book to give me this result. This only thing close I could find was Monotone convergence however we don't know that $\{ f_n \}$ converges. Any help would be appreciated.

Answer 1

Apply the Monotone Convergence Theorem. Define $S_n = \sum_{k=1}^n f_k$. Note that $S_n\nearrow \sum_{k=1}^\infty f_k$ because $f_k$ is non-negative. Then $\int \sum_{k=1}^\infty f_k = \int \lim_n S_n = \lim_n \int S_n = \sum_{k=1}^\infty \int f_k$.