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I would like to show that if $\sum_{n=1}^\infty \int|f_n| d\mu < \infty $ then there must exist a function $f$ with $$f = \sum_{n=1}^\infty f_n$$(or at least has to converge for $x\in A $ with $\mu (A) >0 $). And $$\int fd\mu = \sum_{n=1}^\infty \int f_n d\mu $$ I am sure the $f_n$ are supposed to be measurable.

I am not quite sure how to show this. I have seen many proves that would just say $$\sum_{n=1}^\infty \int|f_n| d\mu = \int\sum_{n=1}^\infty |f_n| d\mu $$ which has to follow from the monotone convergence theorem but I dont quite see how that should work: Let's say I define $g_n= \sum_{j=1}^{n}|f_j|$. Then surely the $g_n$ are increasing, but if I want to use the monotone convergence theorem, there must exist a function $g$ with $g_n \uparrow g$. Isn't this what I have to show in the first place though? If yes, how else would I prove this statement?

Any help would be greatly appreciated

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Set $$ g_n=|\,f_1|+\cdots+|\,f_n|. $$ Then $g_n$ is a sequence of non-negative and increasing and converges to a non-negative measurable function $g$, which is a.e. less the infinity, since, according to Monotone Convergence Theorem $$ \int g=\sum_{n=1}^\infty\int |\,f_n|<\infty. $$ Meanwhile, if $F_n=f_1+\cdots+f_n$, then $F_n$ converges a.e. point-wise to a measurable function $F$ (as $g_n$ also does, and comparison theorem of series applies), and $$ |F_n|\le g, $$ which makes the Lebesgue DCT applicable and $F_n\to F$ is the $L^1$ norm.

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    $\begingroup$ The sequence $g_n$ converges to a measurable function $g$, simply because it is increasing. However, such a limit may attain the value $\infty$. However, the assumption in the OP, guarantees that this can only happen in a set of measure zero. $\endgroup$ – Yiorgos S. Smyrlis Jun 21 '17 at 16:52
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By the MCT $g_n$ converges almost everywhere to an $L^1$ function $g$. Then $\sum_n f_n$ converges absolutely almost everywhere. Using Dominated Convergence with respect to $g$ shows that $\int f=\sum\int f_n$.

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  • $\begingroup$ First of all, thank you very much. I still do not quite see how I can apply the MCT here. The way I know the MCT is: If $g_n$ is increasing and $g_n$ converges to a function $g$ then $\lim_{n \to \infty} \int g_n = \int \lim_{n \to \infty}g_n$ $\endgroup$ – AxiomaticApproach Jun 21 '17 at 16:27
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    $\begingroup$ An increasing sequence of measurable functions will always converge to a measurable function (if you allow $+\infty$ as a value). If the measurable functions are $L^1$ then $\lim\int g_n=\int\lim g_n$. We allow the limit of the integral to be $+\infty$. This is MCT. If $\lim\int g_n<\infty$ then $\lim g_n$ can only be infinite on a set of measure zero, so $(g_n)$ converges almost everywhere. $\endgroup$ – Lord Shark the Unknown Jun 21 '17 at 16:48

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