= Solution
\b[False.] A <typewriter sequence> sweeps successively through finer dyadic partitions. For $k\geq0$ and $j=0,\ldots,2^k-1$, define the <indicator function>
$$
f_{2^k+j}(x)=\mathbf1_{[j2^{-k},(j+1)2^{-k})}(x),\qquad x\in(0,1).
$$
Its <L2 norm> satisfies
$$
\boxed{\|f_{2^k+j}\|_2^2=2^{-k}\longrightarrow0.}
$$
As $n$ 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 $k\geq1$ 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>.
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