# Lebesgue Dominated and Monotone Convergence Theorem Question

Using the Monotone convergence theorem or Lebesgue dominated convergence theorem, prove the following:

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

2. 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

• @Mark But the OP did talk about attempts.
– zhw.
Apr 4, 2020 at 19:25
• OP edited the question. There were no attempts at the beginning. Alright, now I delete my first comment.
– Mark
Apr 4, 2020 at 19:27
• @Mark OK, sorry I handn't seen that.
– zhw.
Apr 4, 2020 at 20:37
• Don't delete your question text. Apr 7, 2020 at 21:30

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 : $$\int_{0}^{1}{f\left(x\right)\mathrm{d}x}\leq 1$$

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.

• 2) Although the problem didn’t specify, I believe E would be measurable, so yes a finite measure
– user767929
Apr 4, 2020 at 20:22
• No, measurable is not the same as finite measure. $[0,\infty)$ is measurable for example.
– zhw.
Apr 4, 2020 at 20:39
• 1) Assuming constant n and increasing x, then wouldn't the integrand be decreasing? Or should I assume constant x and increasing n (then yes it would be increasing). But still, how could I make the assumption from there that $\int$ f dx $\le$ 1?
– user767929
Apr 6, 2020 at 17:37
• Pointwise increasing means for each fixed $x,$ $f_n(x)$ increases as $n$ does.
– zhw.
Apr 6, 2020 at 17:51
• For #2, if m(E) < $\infty$, would the dominating/bounding function be 1/e?
– user767929
Apr 6, 2020 at 19:22