4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What does "integrable" mean? Riemann or Lebesgue? $\endgroup$ – zhw. Dec 14 '18 at 2:33
  • $\begingroup$ Riemann integrability $\endgroup$ – joseph Dec 14 '18 at 2:34
  • 1
    $\begingroup$ So you don't know the dominated convergence theorem? $\endgroup$ – zhw. Dec 14 '18 at 2:35
  • $\begingroup$ I don't know it $\endgroup$ – joseph Dec 14 '18 at 2:38
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Dec 14 '18 at 3:06
6
$\begingroup$

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 $\int_{a} ^{b} f_n(x) \, dx\to 0$.

$\endgroup$
  • $\begingroup$ Very nice answer. $\endgroup$ – RRL Dec 14 '18 at 7:36
1
$\begingroup$

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 $n\rightarrow \infty$ $,\forall x\in [0,1]$.

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.