 Solution

False. A typewriter sequence sweeps successively through finer dyadic partitions. For $k\ge 0$ and $j=0,\dots ,2^k-1$, define the indicator functionIts L2 norm satisfiesAs $n$ tends to infinity so does its block index $k$, proving norm convergence to zero. However, every $x\in (0,1)$ belongs to exactly one of the half-open intervals in each block, so $f_n(x)=1$ infinitely often. For $k\ge 1$ there is also an interval in each block that does not contain $x$, so $f_n(x)=0$ infinitely often. Thus there is no pointwise convergence at any $x\in (0,1)$, let alone convergence almost everywhere.