How to apply Dominated Convergence Theorem to a sequence of functions 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?
 A: You can't, it isn't true.
Take your favorite example of a sequence converging pointwise but not in $L^1$.  For instance, let's take $X = [0,1]$ with Lebesgue measure; then the sequence $n 1_{[0,1/n]}$ will do.  We want this to be $s_n$.  Working backwards, we let $f_n = n 1_{[0,1/n]} - (n-1) 1_{[0, 1/(n-1)]}$ (with $f_1 = 1$).  Then  we have $\sum_{k=1}^\infty f_n(x) = 0$ almost everywhere, and $f=0$ is certainly dominated by a summable (i.e. integrable) function, with $\int f = 0$.  On the other hand, $\int f_1 = 1$ and $\int f_k = 0$ for all $k \ge 2$, so $\sum_{k=1}^\infty \int f_k = 1$.
Generally speaking, having control on the limiting function $f$ is never enough in these situations (and the statement "$f$ is dominated by a summable function" is suspicious because it's the same as just saying "$f$ is summable").  You have to have some sort of uniform control over all the approximating functions $f_n$.
A typical condition under which this does hold is to have either $\sum_{k=1}^\infty  \int |f_n| < \infty$, or $\int \sum_{k=1}^\infty |f_n| < \infty$ (you can use monotone convergence or Tonelli's theorem to see these are equivalent).  In this case you can use either dominated convergence (with dominating function $\sum_{k=1}^\infty |f_n|$) or Fubini's theorem to conclude $\sum_{k=1}^\infty \int f_n = \int \sum_{k=1}^\infty f_n$.
A: 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.
