# Given $f(x)$ is integrable on $[0, 1]$ and $0 < f(x) < 1$, prove that $\int_{0}^{1} (f(x))^{n} \mathop{dx}$ converges to $0$.

Given $$f(x)$$ is integrable on $$[0, 1]$$ and $$0 < f(x) < 1$$, prove that $$\int_{0}^{1} (f(x))^{n} \mathop{dx}$$ converges to $$0$$.

I understand why the statement is true intuitively because as $$n \to \infty$$, since $$f$$ lies between $$0$$ and $$1$$, it will be like a fractional value, which converges to $$0$$ since the fractions get smaller and smaller.

However, I am not sure about how to prove this rigorously.

• What does "integrable" mean? Riemann or Lebesgue? – zhw. Dec 14 '18 at 2:33
• Riemann integrability – joseph Dec 14 '18 at 2:34
• So you don't know the dominated convergence theorem? – zhw. Dec 14 '18 at 2:35
• I don't know it – joseph Dec 14 '18 at 2:38
• Without dominated convergence it appears a bit difficult. The sequence of integrals is decreasing and positive so the limit exists. We need some sort of contradiction to prove that the limit can't be positive. – Paramanand Singh Dec 14 '18 at 3:06

Here is a proof which assumes some amount of measure theory (and I think this is unavoidable, but I may be wrong in thinking so).

Let $$f_n(x) =(f(x)) ^n$$ then each $$f_n(x)$$ is Riemann integrable on $$[0,1]$$ and hence the set $$D_n$$ of its discontinuities is of measure $$0$$ and thus the set $$D=\bigcup\limits_{n=1}^{\infty}D_n$$ is of measure $$0$$. Let $$\epsilon>0$$ be given. Then there is a sequence of open intervals $$\{J_n\}$$ such that $$D\subseteq \bigcup\limits_{n=1}^{\infty} J_n$$ and the length of these intervals $$J_n$$ combined is less than $$\epsilon$$.

Next $$f_n(x) \to 0$$ as $$n\to\infty$$ for all $$x\in[0,1]$$. Let $$x\in[0,1]\setminus D$$. Then we have a positive integer $$n_x$$ depending on $$x$$ such that $$f_n(x) <\epsilon$$ for all $$n\geq n_x$$. By continuity of $$f_{n_x}$$ at $$x$$ it follows that there is a neighborhood $$I_x$$ such that $$f_{n_x} (x) <\epsilon$$ for all $$x\in I_x$$. Since $$f_n$$ is decreasing it follows that we have $$f_n(x) <\epsilon$$ for all $$x\in I_x$$ and all $$n\geq n_x$$.

Now the set of all neighborhoods $$I_x$$ as $$x$$ varies in $$[0,1]\setminus D$$ together with the intervals $$J_n$$ forms an open cover for $$[0,1]$$ and thus by Heine Borel theorem a finite number of these intervals covers $$[0,1]$$. Thus we have $$[0,1]\subseteq \bigcup\limits_{i=1}^{p}I_{x_i} \cup\bigcup\limits_{i=1}^{q}J_i$$ Let $$N$$ be the maximum of integers $$n_{x_1},n_{x_2},\dots,n_{x_p}$$ then we have $$f_n(x) <\epsilon, \forall x\in\bigcup\limits _{i=1}^{p}I_{x_i} , \forall n\geq N$$ The end points of $$J_1,J_2,\dots,J_q$$ which lie in $$[0,1]$$ partition it into a finite number of subintervals. Denote the union of all those subintervals which contain points of $$J_1,\dots, J_q$$ as $$A$$ and let the union of remaining subintervals be denoted by $$B$$. Then length of $$A$$ is less than $$\epsilon$$ and $$f_n(x) <\epsilon$$ for all $$n\geq N$$ and all $$x\in B$$. Thus we have $$\int_{0}^{1}f_n(x)\,dx=\int_{A}f_n(x)\,dx+\int_{B}f_n(x)\,dx<\epsilon +\epsilon =2\epsilon$$ for all $$n\geq N$$. Therefore $$\int_{0}^{1}f_n(x)\,dx\to 0$$ as $$n\to \infty$$.

Note that the above argument actually proves the following result:

Theorem: Let $$\{f_n\}$$ be a sequence of functions $$f_n:[a, b] \to\mathbb {R}$$ such that each $$f_n$$ is non-negative and Riemann integrable on $$[a, b]$$ and $$f_n(x) \geq f_{n+1}(x),\forall x\in[a, b]$$ and $$f_n(x) \to 0$$ point wise almost everywhere in $$[a, b]$$ then $$\int_{a} ^{b} f_n(x) \, dx\to 0$$.

• Very nice answer. – RRL Dec 14 '18 at 7:36

You may use the following theorem due to Arzelà :---

Let $$\{f_n\}$$ be a sequence of Riemann integrable Functions on $$[a,b]$$ and converges point-wise to $$f$$, also there is a positive number $$M$$ such that $$|f_n(x)|≤M,\forall x\in [a,b],\forall n\in \Bbb N$$. Now if $$f$$ is Riemann integrable over $$[a,b]$$ then , $$\lim_{n\rightarrow \infty}\int_a^bf_n(x)dx=\int_a^b\lim_{n\rightarrow \infty} f_n(x)dx=\int_a^b f(x) dx.$$

Here $$f_n(x)=(f(x))^n\rightarrow 0$$ as $$n\rightarrow \infty$$ $$,\forall x\in [0,1]$$.

• +1 Proving Arzela theorem is difficult. I hope the asker is allowed to use Arzela theorem for the practice exercise. – Paramanand Singh Dec 14 '18 at 5:23

Since $$f$$ is integrable, it is measurable. By Lusin's theorem, for any $$\varepsilon>0$$ there exists a compact set $$K\subset [0,1]$$ such that $$f$$ is uniformly continuous on $$K$$ and $$|K|>1-\varepsilon$$. Uniform continuity implies that $$\sup_{x\in K} f(x) = \lambda<1$$. Thus \begin{align} \int_{[0,1]} f(x)^n\, dx &= \int_{K} f(x)^n\, dx + \int_{[0,1]\backslash K} f(x)^n\, dx \\ &\le |K|\lambda^n + \varepsilon\cdot1. \end{align} Take limit as $$n\to\infty$$ yields $$\limsup_{n\to \infty} \int_{[0,1]} f(x)^n\, dx \le \varepsilon.$$ Since the above hold for any $$\varepsilon>0$$, we have $$\int_{[0,1]} f(x)^n\, dx\to 0$$ as wanted.