= Solution
{c}
By part <3/b/i>, $\chi_{E_n}|g|\to0$ <almost everywhere>, while $0\leq\chi_{E_n}|g|\leq|g|$ and $g$ is <Lebesgue integrable>. The <dominated convergence theorem> gives
$$
\boxed{\int_{E_n}|g(x)|\,dx\longrightarrow0.}
$$
To deduce <absolute continuity of the Lebesgue integral>, suppose the stated uniform conclusion failed for some $\varepsilon_0>0$. For each $n$ one could choose a <measurable set> $A_n$ with $m(A_n)<2^{-n}$ but $\int_{A_n}|g|\geq\varepsilon_0$. Applying the result just proved to this <sequence> contradicts the lower bound. Hence
$$
\boxed{\forall\varepsilon>0\ \exists\delta>0:\quad m(A)<\delta\Longrightarrow\int_A|g|<\varepsilon}
$$
for every <measurable set> $A$.
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