If $\sum^\infty_{n=1} \int |f_n| d\mu<\infty$, then $\sum_{n=1}^\infty f_n(x) d\mu$ converges to $f(x)$ , where $f_n$ and $f$ are integrable?

Question

Want to prove that if $\{f_n\}$ is a sequence of integrable functions and $\sum^\infty_{n=1} \int |f_n| d\mu<\infty$, then $\sum_{n=1}^\infty f_n(x) d\mu$ is convergent to integrable function $f(x)$, and also that $\int f d \mu = \sum^\infty_{n=1}\int f_n d \mu$.

I considered the inequalities

$\int \sum^\infty_{n=1}f_n d\mu = \sum^\infty_{n=1} \int f_n d\mu \leq|\sum^\infty_{n=1} \int f_n d\mu|\leq\sum^\infty_{n=1} \int |f_n| d\mu<\infty$,

but I'm not sure how to proceed e.g. to show that the integrand is finite as well. Also, since the question is asked in the context of the Lebesgue monotone convergence theorem i considered defining

$g_m(x) = \sum^m_{n=1}|f_n|$,

where $g_m$ is integrable since it can be shown that $f_n$ being integrable implies that $|f_n|$ is integrable and because finite sums of integrable functions are integrable. Because $\{g_m\}$ is a monotone increasing sequence of non-negative integrable functions we have by the Lebesgue monotone convergence theorem that

$\lim_{m \rightarrow \infty} \int g_m d \mu = \int \sum^\infty_{n=1}|f_n| d \mu < \infty $,

but didn't managed to use this to make a clear argument for the assertion. Could someone please help?

Answer 1

Let $S_{N}=\sum_{n=1}^{N}f_{n}$, we are to show that $\{S_{N}\}$ is Cauchy in $L^{1}$, here is it:

For $N>M$, then $\|S_{N}-S_{M}\|_{L^{1}}=\int|S_{N}-S_{M}|=\int|\sum_{n=M+1}^{N}f_{n}|\leq\int\sum_{n=M+1}^{N}|f_{n}|=\sum_{n=M+1}^{N}\int|f_{n}|$.

Note that $\sum_{n=1}^{\infty}\int|f_{n}|<\infty$ entails that $\sum_{n=M+1}^{N}\int|f_{n}|$ is arbitrary small for large $N,M$, this is the Cauchy criterion of convergent sequence of real numbers, this entails that $\{S_{N}\}$ is convergent in $L^{1}$, say, $\|S_{N}-f\|_{L^{1}}\rightarrow 0$ for some $f\in L^{1}$.

On the other hand, $\sum_{n=1}^{\infty}\int|f_{n}|=\int\sum_{n=1}^{\infty}|f_{n}|$ and so $\int\sum_{n=1}^{\infty}|f_{n}|<\infty$, this entails $\sum_{n=1}^{\infty}|f_{n}(x)|<\infty$ a.e. and hence $\sum_{n=1}^{\infty}f_{n}(x)$ exists a.e.

Going back to that $\|S_{N}-f\|_{L^{1}}\rightarrow 0$, a standard fact says that $S_{N_{k}}(x)\rightarrow f(x)$ a.e. for a subsequence $\{S_{N_{k}}\}$, that is, $\sum_{n=1}^{N_{k}}f_{n}(x)\rightarrow f(x)$ a.e. as $k\rightarrow\infty$.

But we already had $\sum_{n=1}^{\infty}f_{n}(x)$ exists a.e., this entails $f(x)=\sum_{n=1}^{\infty}f_{n}(x)$ a.e.

Answer 2

A shorter answer:

As usual, $\int\sum_{n=1}^{\infty}|f_{n}|<\infty$ entails $\sum_{n=1}^{\infty}f_{n}(x)$ exists a.e.

We have \begin{align*} \int\left|\sum_{n=1}^{N}f_{n}(x)-\sum_{n=1}^{\infty}f_{n}(x)\right|&=\int\left|\sum_{n=N+1}^{\infty}f_{n}(x)\right|\\ &\leq\int\sum_{n=N+1}^{\infty}|f_{n}(x)|\\ &=\sum_{n=N+1}^{\infty}\int|f_{n}(x)|\\ &=\sum_{n=1}^{\infty}\int|f_{n}|-\sum_{n=1}^{N}\int|f_{n}|\\ &\rightarrow 0, \end{align*} so $\sum_{n=1}^{N}f_{n}$ converges to the function $\sum_{n=1}^{\infty}f_{n}$ in $L^{1}$, of course, $\sum_{n=1}^{\infty}f_{n}$ is necessary in $L^{1}$ since \begin{align*} \int\left|\sum_{n=1}^{\infty}f_{n}(x)\right|\leq\int\left|\sum_{n=1}^{N}f_{n}(x)-\sum_{n=1}^{\infty}f_{n}(x)\right|+\int\left|\sum_{n=1}^{N}f_{n}(x)\right|<\infty. \end{align*}