Let $f_m:[0,1]\mapsto\mathbb{R}$ be defined as for $m\in\mathbb{N}$ such that $\exists m = 2^n + k,\ n\in\mathbb{N},\ k\in\{0,1,2,\dots,2^n-1\}$ $$ f_{2^n+k}(x) = \begin{cases} 1 & x\in[\frac{k}{2^n},\frac{k+1}{2^n}]\\ 0 & \text{otherwise} \end{cases} $$
Prove that $\{f_n\}$ converges to some $f$ in mean, find $f$ explicitly and then show that $\{f_n\}$ diverges pointwise.
By "converges to $f$ in mean," we use the following definition:
A sequence of functions $\{f_n\}$, with $f_n:[a,b]\mapsto\mathbb{R}$, converges in mean to a function $f$ if and only if $$ \lim\limits_{n\rightarrow\infty}\left(\int_a^b [f_n(x) - f]^2 dx\right)^{\frac{1}{2}} = 0 $$
For the mean convergence, follow Robert Israel's hint.
For the pointwise divergence : what can you say about $(f_{2^n+2^n-1}(1))_n$ and $(f_{2^n}(1))_n$ ? and what can you say about a sequence which has at least two subsequences that converge to a different value ?