= Solution
{c}
The points belonging to infinitely many $E_n$ form their <limit superior of sets>:
$$
G=\bigcap_{N\geq1}\bigcup_{n\geq N}E_n.
$$
For every $N$, <countable subadditivity> gives
$$
0\leq m(G)\leq\sum_{n\geq N}m(E_n)\leq\sum_{n\geq N}2^{-n}=2^{1-N}\longrightarrow0.
$$
Thus $\boxed{m(G)=0}$. Outside this <null set>, each point belongs to only finitely many $E_n$, so its <indicator functions> are eventually zero. Therefore $\boxed{\chi_{E_n}(x)\to0\text{ almost everywhere}}$. This is the summable-measure argument behind the first <Borel-Cantelli lemma>; no <independence> assumption is needed.
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