Term by term integration. How do I go about proving this statement? 
If $$\sum_{k=1}^\infty f_k(x)$$is a series of nonnegative measurable functions and $$\sum_{k=1}^\infty \left(\int_Ef_k(x)dx\right)$$ converges, then $$\sum_
{k=1}^\infty f_k(x)$$ converges almost everywhere and $$\int_E\left(\sum_{k=1}^\infty f_k(x)\right)dx=\sum_{k=1}^\infty\left(\int_E f_k(x)dx\right)$$

We have for each $x \in E$ the convergence of partial sums as $n \to \infty$ according to
$$\sum_{k=1}^n f_k(x) \uparrow \sum_{k=1}^\infty f_k(x) \leqslant +\infty$$

Why is it true that the sequence of partial sums is non-decreasing and
  must always converge to a possibly extended nonnegative real number?

By the monotone convergence theorem,
$$\sum_{k=1}^\infty\int_E f_k(x) \, dx = \lim_{n \to \infty}\sum_{k=1}^n \int_E f_k(x) \, dx = \lim_{n \to \infty} \int_E \sum_{k=1}^nf_k(x) \, dx = \int_E \sum_{k=1}^\infty f_k(x)\, dx$$
We are given that the series on the LHS converges, and it follows that
$$\int_E \sum_{k=1}^\infty f_k(x)\, dx < +\infty$$

What does this tell you about $ \sum_{k=1}^\infty f_k(x)$?

