TITLE: How to apply Dominated Convergence Theorem to a sequence of functions
QUESTION [5 upvotes]: Let $(X,\Sigma, u)$ be a measure space. Suppose $(f_n)$ is a sequence of measurable functions $f(x) = \sum_{n=1}^{\infty} f_n(x)$ converges $u$ almost everywhere and $|f(x)|\leq g$ for some $u$-summable function $g$.
I am trying to show that $\int_X f du = \sum_{n = 1}^{\infty} \int_X f_n du$.
I am pretty sure we must use the Lebesgue Dominated Convergence Theorem. We know that $\int_X |f| du \leq \int_X g du$ and $f$ is measurable as it is the $u$-almost everywhere pointwise limit of measurable functions. Hence $f$ is $u$-summable.
I want to consider the sequence $s_n= \sum_{k = 1}^{n} f_n $ since then $s_n\rightarrow f$. Now if we can show that $|s_n|<h$ for some $u$-summable function $h$, then the result will follow immediate from the dominated convergence theorem. 
However I am struggling to show that $|s_n|$ can be dominated (of course things would be much easier if $f_n$ were positive). How can I show this?

REPLY [2 votes]: This is wrong in general. Here is a counter-example. I will take $(X,\Sigma,\mu)=(\mathbb{R}, B_{\mathbb{R}},\lambda)$, and my sequeence of functions is composed of constant functions: $f_0(x)\equiv1$, and for $n\ge 1$, $f_n(x)\equiv \frac{1}{n+1}-\frac1n$.
The series $\sum_{n=0}^\infty f_n(x)$ converges every where to the constant function $f(x)\equiv0$, which is clearly integrable (one may take $g=f=0$). But clearly the equality
$\sum_{n=0}^\infty \int_{\mathbb{R}}f_nd\lambda=0$ makes no sense.
Thus, there is a missing suplementary condition for the proposed equality to hold.