= Solution
{c}
The a.e. limit $f$ is nonnegative. For a fixed <measurable set> $A$, <Fatou's lemma> gives $\int_Af\leq\liminf_n\int_Af_n$. Apply the same lemma to $A^c$ and use convergence of the total integrals:
$$
\limsup_n\int_Af_n
=\int_{\mathbb R}f-\liminf_n\int_{A^c}f_n
\leq\int_{\mathbb R}f-\int_{A^c}f=\int_Af.
$$
The lower and upper bounds prove $\boxed{\int_Af_n\to\int_Af}$.
For the asserted <L1 convergence>, choose an <integer> $k$ so large that $\int_{|x|>k}f<\varepsilon$, possible by <Lebesgue integrability>. The first result applied to this fixed tail <set> makes $\int_{|x|>k}f_n<2\varepsilon$ eventually. The additional local hypothesis makes $\int_{-k}^k|f_n-f|<\varepsilon$ eventually. Therefore
$$
\int_{\mathbb R}|f_n-f|
\leq\int_{-k}^k|f_n-f|+\int_{|x|>k}f_n+\int_{|x|>k}f<4\varepsilon.
$$
Thus $\boxed{\|f_n-f\|_{L^1(\mathbb R)}\to0}$.
In fact the extra local hypothesis is unnecessary here. <Scheffé's lemma> follows directly from $\min\{f_n,f\}\to f$ <almost everywhere> and domination by $f$:
$$
\|f_n-f\|_1=\int f_n+\int f-2\int\min\{f_n,f\}\longrightarrow0.
$$
This stronger argument uses precisely the already assumed nonnegativity, a.e. convergence, and convergence of total integrals.
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