A problem on interchange of limit and integration Suppose $\lim_{h\to 0}f_n(h) = 0$ such that $g(h)=\sum_{n=1}^{\infty}f_n(h)$ converges for any $h$. Can we tell that $\lim_{h\to0}g(h) = 0$ ? i.e can we change the order of limit and sum here ? If not what is needed to make it happen ? 
I don't know how to use DCT/BCT here.
Is it true if $f_n(h) = \int_{E_n} |f(x+h) -f(x)| dx$ where $E_n$ is an interval of length $[nh,(n+1)h]$ where $f$ is bounded and integrable. 
 A: No: Consider $f_n(h)=h^{n-1}$ for $h\in(-1,1),n\in\mathbb N$. Then $f_n(h)\to0$ for $h\to0$ and any $n$, but
$$
g(h)=\sum_{n\in\mathbb N}f_n(h)=\sum_{n\in\mathbb N}h^{n-1}=\frac{1}{1-h},
$$
which does not tend to 0 as $h\to0$.
To interchange limits you need that the series converges uniformly
EDIT: This example doesnt work (see comment). But how about this:
Since $\mathbb Q\backslash\{0\}$ is countable, there is a bijective mapping $a:\mathbb N\to\mathbb Q\backslash\{0\}$. Now define $f_n(h)=0$ if $h\neq a(n)$ and $f_n(h)=1$ if $h=a(n)$. Then each $f_n$ is continuous at $0$, since $f_n$ is 0 everywhere except at $a(n)\neq0$. Since $a$ is bijective we have that $g(h)=1$ if $h\in\mathbb Q\backslash\{0\}$ and $g(h)=0$ otherwise. In particular, $g(1/n)=1$ for any $n\in\mathbb N$.
A: $$\lim_{h\to0}g(h) = \lim_{h \to 0} \sum_{n=1}^{\infty} f_n(h)=\sum_{n=1}^{\infty} \left({\lim_{h \to 0} f_n(h)}\right) = 0 $$
as $$\sum_{n=1}^{\infty} \left({\lim_{h \to 0} f_n(h)}\right) = \sum_{n=1}^{\infty} 0 = 0 $$
