Lebesgue Dominated and Monotone Convergence Theorem Question Using the Monotone convergence theorem or Lebesgue dominated convergence theorem, prove the following:

*

*Let $f\ge0$ and $\int_0^1\frac{f(x)}{n+x} \,dx \le\frac{1}{n}$ for all $n.$ Show that $\int_0^1 f \,dx\le1$


*If $f_n\ge 0$ and $f_n \to f$ on $E$ (a.e. = almost everywhere). Then $$\lim_{n \to \infty}\int_E f_n e^{-fn} \,dx = \int_E fe^{-f} \, dx$$
For number 1, I believe it is best to use the monotone convergence theorem, but I am not sure how to prove the functions are even monotone, nor how to get rid of the n+x denominator
For number 2, it worked if I assumed the functions were monotone, but I was told it was wrong to straight up assume that (I can only make assumptions based on what is given in the problem). So the Lebesgue dominated convergence theorem would be more useful here, but I am not sure how to create an upper bound on these functions
 A: Note that : $$ \left|n\int_{0}^{1}{\frac{f\left(x\right)}{n+x}\,\mathrm{d}x}-\int_{0}^{1}{f\left(x\right)\mathrm{d}x}\right|=\int_{0}^{1}{\frac{x f\left(x\right)}{n+x}\,\mathrm{d}x}\leq\frac{1}{n}\int_{0}^{1}{xf\left(x\right)\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}0 $$
Thus, $$ \lim_{n\to +\infty}{n\int_{0}^{1}{\frac{f\left(x\right)}{n+x}\,\mathrm{d}x}}=\int_{0}^{1}{f\left(x\right)\mathrm{d}x} $$
Now we have that $ \left(\forall n\in\mathbb{N}\right),\ n\int\limits_{0}^{1}{\frac{f\left(x\right)}{n+x}\,\mathrm{d}x}\leq 1 $, Taking limits both sides of the inequality gives : $$ \int_{0}^{1}{f\left(x\right)\mathrm{d}x}\leq 1 $$
A: The hypotheses imply
$$\tag 1 \int_0^1\frac{n}{n+x}f(x)\,dx\le 1$$
for all $n.$ Define $f_n(x) = \dfrac{n}{n+x}f(x).$ Then $f_1 \le f_2 \le \cdots$ and $f_n\to f$ pointwise everywhere. By the MCT, $\int_0^1f_n \to \int_0^1 f.$ Since each $\int_0^1f_n\le 1,$$\int_0^1 f\le 1$ as well.
2) This fails if $m(E)=\infty.$ Example: Let $E=[0,\infty).$ For $n=1,2,\dots$ let $f_n=\chi_{[n,n+1]}.$ Then $f_n \to f\equiv 0$ pointwise everywhere on $E.$ But
$$\int_E f_ne^{-f_n} = \int_n^{n+1} \frac{1}{e}\,dx = \frac{1}{e}$$
for all $n,$ yet $\int_E fe^{-f} = \int_E 0 =0.$ 
The result holds if $m(E)<\infty.$ Hint: DCT.
