Pointwise Convergence implies convergence of integral of Exponential

Question

$f_n \to f$ pointwise, where $f_n$ and f are measurable and $f_n(x) \geq 0$ a.e.

Show $\int_{[0,1]}f_n e^{-f_n}dm \to \int_{[0,1]}f e^{-f}dm$.

Note that the following is not necessarily true: $\int_{[0,1]}f_n dm \to \int_{[0,1]}f dm$.

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My attempt:

$|\int_{[0,1]}f_n e^{-f_n}dm - \int_{[0,1]}f e^{-f}dm| =|\int_{[0,1]}[(f_n e^{-f_n} -f_ne^{-f})+(f_ne^{-f} -f e^{-f})]dm|$

$ =|\int_{[0,1]}[f_n (e^{-f_n} -e^{-f})+e^{-f}(f_n -f )]dm|$

But just because $f_n \to f$ a.e. does not mean that $\int fn \to \int f$. I'm not sure where to go from here. Thanks! I noticed that I did not use the nonnegativity, but I'm not sure where that will be useful.

Answer 1

Since $f_n \geq 0$, \begin{equation} 0\leq f_ne^{-f_n} \leq 1/e \end{equation} and use Dominated Convergence with constant function 1/e, which is integrable on $[0,1]$.

Let $f_n = n\mathbb{1}_{[0,1/n)}$, which converges 0 pointwise, but \begin{equation} 1=\int_{[0,1]}f_ndm \neq \int_{[0,1]}0dm = 0 \end{equation}