Example where $\lim\limits_{m\rightarrow\infty} \int_E f_m =0$ and $f_m(x) \not\rightarrow 0$ I am looking for an example of a sequences of non-negative and measurables functions with $\lim\limits_{m\rightarrow\infty} \int_E f_m =0$ and $f_m \nrightarrow 0 \:\:\forall\:x\in E$
 A: Hint: Consider $E=[0,1]$ and $$\begin{align*} f_1(x) &:= 1_{[0,1/2]}(x) \\ f_2(x) &:= 1_{[1/2,1]}(x) \\ f_3(x) &:= 1_{[0,1/4]}(x) \\ f_4(x) &:= 1_{[1/4,1/2]}(x) \\ \vdots & \end{align*}$$
A: On $E=[0,1]$, take $f_m$ to equal the indicator function on $[s_m,s_{m+1}]$, where $s_m=\sum_{n=1}^m\frac{1}{n}$ modulo $1$.  (If $s_m>s_{m+1}$, then instead use $[0,s_{m+1}]\cup[s_m,1]$.)
A: Here's an easy one: just let $f_m = \mathbb{1_Q}$ for all $m$. $f$ is clearly not the zero function, but as a simple function it has integral equal to $\mu(\mathbb{Q})=0$.
A: The usual example I've seen here is the "moving block". Rather than writing a precise definition, let me describe the definition in words. 
At stage $n=0,1,\dots,$ we will append $2^n$ functions to our sequence, specifically the indicator functions of $[k 2^{-n},(k+1)2^{-n}]$ as $k$ ranges from $0$ to $2^n-1$. So the first function is $1_{[0,1]}$, the next two are indicator functions of halves of the interval, the next four are indicator functions of quarters of the interval, etc.
Then the integral of each of the $2^n$ functions from stage $n$ is $2^{-n}$, which converges to zero. But pointwise convergence occurs nowhere, because any point is represented in one of the $2^n$ functions being listed at stage $n$.
