Proof of properties of $L^1(\mu, X)$ Hi
I'm trying to prove some properties of $L^1$, the space of $\mu$-integrable functions.
I have $$ f_n \in L^1$$ and $$ \sum_{n \geq 1} || f_n ||_{1} \lt \infty$$ and I used it to prove $$ \sum_{n\geq 1} f_n(x) \lt \infty $$ $\mu$-almost-everywhere.
The next thing I would like to show is that $$ \sum_{n\geq 1} f_n(x) $$ is measurable. 
I know how to show that the pointwise (i.e. $\forall x \in X$) limit of $\mu$-measurable functions is measurable, but I don't know how to cope with a sequence that only converges $\mu$-almost everywhere, i.e. if the limit is not even defined in some points inside a $\mu$-negligible set.
Many thanks for your help!
 A: denote by $A$ the set where $\sum f_n$ doesn't converge, then you showed that $\mu(A)=0$. Let $g_n = f_n \chi_{X-A}$, then $X-A$ is a measurable set, hence $g_n$ are measurable functions.
$\sum g_n = \sum f_n$ almost everywhere, by the definition of $g_n$, and $\sum_1^m g_n$ is pointwise convergent, so its limit is measurable. now change it back on the set $A$ to be $\sum f_n$, to get that $\sum f_n$ is also measurable (Since we changed a measurable function on a zero measure set). In other words $h = \sum g_n - \sum f_n$ is almost everywhere zero, so it is measurable, hence $\sum f_n = \sum g_n -h$ is measurable.

Suppose f is 0 almost everywhere, and let $A=f^{-1}(0)$ so $\mu(X-A)=0$, then f is measurable. For every $t\in \mathbb{R}$ we have $B=f^{-1}(t,\infty)$ satisfies $A\subseteq B$ or $B \subseteq X-A$. In the first case you have $\mu(X-B)\leq \mu (X-A)=0$ so $B-X$ is measurable (with measure zero) hence B is measurable. In the second case, $\mu (B) \leq \mu(X-A)=0$ and again B is measurable. (actually, we need to do this with the outer measure of B, X-B).
any way, now that you know that B is measurable, then f is measurable (one of the equivalent definitions). This shows that a function that is zero almost everywhere is measurable, so you can always add it to another measurable function to get another measurable function, or in other words, you can always change a measurable function in a zero measure set, and still have a measurable function.
(regarding the answer in your other question, I assumed that the measure is complete - meaning that if $A \subseteq B$ and $\mu(B)=0$ then A is measurable and then its measure must be zero. This is true for example in the Lebesgue measure)
