If the integral of a series of functions converges to zero, does that series also converge pointwise to zero? Suppose I have the condition $\int |f_n| \rightarrow 0$. Does this give me the condition that $f_n \rightarrow 0$ pointwise? It seems to be true, except that if you redefine up to countably many points of the $f_n$, it will not change the integral, but it will prevent pointwise convergence at those points. Maybe almost everywhere convergence is what you get? If there is a theorem relating to this, I would find it very helpful.
 A: There is the following standard example. Let $H_n = \sum_1^n \frac1n$. Define the closed sets$$A_n = [H_n, H_{n+1}] \mod 1 \subset [0,1],$$ where we always choose $|A_n| = \frac1{n+1}$ by treating $[0,1]$ as the periodic circle (so $A_1 = [0,1/2]$, $A_2 = [1/2,5/6]$, $A_3=[5/6,1]\cup[0,1/12],\dots$). 
Define
$$ f(x) = \mathbb 1_{x\in A_n}.$$
Note $f\to 0$ in $L^1$. For every fixed $x$, $f(x) = 1$ infinitely often; so $f$ does not converge pointwise or almost everywhere. What we can say is $f$ converges in measure (to the zero function.) Also, as mentioned in comments, there is a.e. convergence along a subsequence.
A: No. Take, for example, $f_n(x)=x^n$ on $[0,1]$.
Then, $$\int_0^1 |f_n(x)|dx = \int_0^1x^n dx = \frac{1}{n+1} \to 0$$ but $f_n$ does not converge to the zero function.
A: Almost everywhere convergence is also not true. Here is a well known example:
Consider the intervals $[\frac {i-1} {2^{n}},\frac {i} {2^{n}})$, $1 \leq i \leq 2^{n}, n \geq 1$. Arrange the indicator functions of these intervals in a sequence $f_1,f_2,...$. Then $\int |f_n| \to 0$ but $f_n(x)$ does not tend to $0$ for any $x \in (0,1)$. 
Two conclusions you can draw are: $f_n$ tends to $0$ in measure and there is a subsequence which converges almost everywhere. 
