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It is rather straightforward to show that the sum of two measurable functions is also measurable. Therefore we can extend the logic to say that $\sum\limits_{i=1}^n f_i$ is measurable providing $f_i$ is measurable. However can we take $n\rightarrow\infty$ and say that $\sum\limits_{i=1}^\infty f_i$ is a measurable function?

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    $\begingroup$ Pointwise limits of measurable functions are measurable. Apply that to the partial sums. $\endgroup$ – Daniel Fischer Aug 1 '13 at 15:59
  • $\begingroup$ Functions from where to where? And by sum of functions, do you mean like $(f+g)(x) = f(x) + g(x)$ or direct sum of functions on the direct sum of the domains? $\endgroup$ – Daniel Robert-Nicoud Aug 1 '13 at 16:00
  • $\begingroup$ I am not sure if your first sentence is so straight forward; I would prefer to have the math for this and know about what type of function etc. you are talking? $\endgroup$ – al-Hwarizmi Aug 1 '13 at 16:02
  • $\begingroup$ Functions from an arbitrary set to the extended real-domain and I mean $(f+g)(x)=f(x)+g(x)$. $\endgroup$ – Simon Aug 1 '13 at 16:03
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As Daniel pointed out, the pointwise limit of measurable functions is measurable.

If we define $g_n=\sum_{i=1}^n f_i$, then as long as $\sum_{i=1}^{\infty} f_i = \lim_{n\to\infty}g_n$ converges, it defines a measurable function.

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  • $\begingroup$ It should be fine for the partial sums to converge to a value in the extended reals i.e. $\infty$ or $-\infty$, is that right? In which case the pointwise limit is the constant function $x \mapsto \infty$. $\endgroup$ – jII Oct 22 '18 at 0:04
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You need some assumptions here, otherwise your infinite sum will not converge, think of $f_n(x)=1$. See https://en.wikipedia.org/wiki/Dominated_convergence_theorem for details.

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  • $\begingroup$ In your example is it not the case that the series converges to the constant function $x\mapsto\infty$? In any case dominated convergence seems like overkill here, instead of e.g. $\forall n:f_n\geq0$ (and monotone convergence theorem). $\endgroup$ – Alp Uzman Mar 10 '16 at 7:50
  • $\begingroup$ This answer does not seem correct to me. The dominated convergence theorem relates to interchanging an integral and a limit. The question is about whether the pointwise limit of a sequence of measurable functions is measurable and not whether it is integrable. Maybe I'm missing something about @MoisheCohen's answer. $\endgroup$ – jII Oct 21 '18 at 23:57

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