# Is the countable sum of a set of measurable functions also measurable?

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?

• Pointwise limits of measurable functions are measurable. Apply that to the partial sums. – Daniel Fischer Aug 1 '13 at 15:59
• 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? – Daniel Robert-Nicoud Aug 1 '13 at 16:00
• 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? – al-Hwarizmi Aug 1 '13 at 16:02
• Functions from an arbitrary set to the extended real-domain and I mean $(f+g)(x)=f(x)+g(x)$. – Simon Aug 1 '13 at 16:03

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.
• 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 $x \mapsto \infty$. – jII Oct 22 '18 at 0:04
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.
• 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). – Alp Uzman Mar 10 '16 at 7:50