 Solution

The a.e. limit $f$ is nonnegative. For a fixed measurable set $A$, Fatou's lemma gives ${\int }_{A}f\le {\mathrm{liminf}}_n{\int }_A{f}_n$. Apply the same lemma to $A^c$ and use convergence of the total integrals:The lower and upper bounds prove $\boxed{{\int }_A{f}_n\to {\int }_{A}f}$.

For the asserted L1 convergence, choose an integer $k$ so large that ${\int }_{|x|>k}f<\epsilon$, possible by Lebesgue integrability. The first result applied to this fixed tail set makes ${\int }_{|x|>k}{f}_n<2\epsilon$ eventually. The additional local hypothesis makes ${\int }_{-k}^k|f_n-f|<\epsilon$ eventually. ThereforeThus $\boxed{\Vert f_n-f{\Vert }_{L^1(\mathbb{R})}\to 0}$.

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 $f$:This stronger argument uses precisely the already assumed nonnegativity, a.e. convergence, and convergence of total integrals.