Vrftsjtryk

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.

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 $D=bigcuplimits_{n=1}^{infty}D_n$ is of measure $0$. Let $epsilon>0$ be given. Then there is a sequence of open intervals ${J_n}$ such that $Dsubseteq bigcuplimits_{n=1}^{infty} J_n$ and the length of these intervals $J_n$ combined is less than $epsilon$.

Next $f_n(x) to 0$ as $ntoinfty $ for all $xin[0,1]$. Let $xin[0,1]setminus D$. Then we have a positive integer $n_x$ depending on $x$ such that $f_n(x) <epsilon$ for all $ngeq n_x$. 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 $xin I_x$. Since $f_n$ is decreasing it follows that we have $f_n(x) <epsilon$ for all $xin I_x$ and all $ngeq n_x$.

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 $$[0,1]subseteq bigcuplimits_{i=1}^{p}I_{x_i} cupbigcuplimits_{i=1}^{q}J_i$$ Let $N$ be the maximum of integers $n_{x_1},n_{x_2},dots,n_{x_p}$ then we have $$f_n(x) <epsilon, forall xinbigcuplimits _{i=1}^{p}I_{x_i} , forall ngeq N$$ 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 $ngeq N$ and all $xin B$. Thus we have $$int_{0}^{1}f_n(x),dx=int_{A}f_n(x),dx+int_{B}f_n(x),dx<epsilon +epsilon =2epsilon $$ for all $ngeq N$. Therefore $int_{0}^{1}f_n(x),dxto 0$ as $nto infty $.

Note that the above argument actually proves the following result:

Theorem: Let ${f_n} $ be a sequence of functions $f_n:[a, b] tomathbb {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 xin[a, b] $ and $f_n(x) to 0$ point wise almost everywhere in $[a, b] $ then $int_{a} ^{b} f_n(x) , dxto 0$.

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 $|f_n(x)|≤M,forall xin [a,b],forall nin Bbb N$. Now if $f$ is Riemann integrable over $[a,b]$ then , $$lim_{nrightarrow infty}int_a^bf_n(x)dx=int_a^blim_{nrightarrow infty} f_n(x)dx=int_a^b f(x) dx.$$

Here $f_n(x)=(f(x))^nrightarrow 0$ as $nrightarrow infty$ $,forall
xin [0,1]$.

Since $f$ is integrable, it is measurable. By Lusin's theorem, for any $varepsilon>0$ there exists a compact set $Ksubset [0,1]$ such that $f$ is uniformly continuous on $K$ and $|K|>1-varepsilon$. Uniform continuity implies that $sup_{xin K} f(x) = lambda<1$. Thus
$$begin{align}
int_{[0,1]} f(x)^n, dx &= int_{K} f(x)^n, dx + int_{[0,1]backslash K} f(x)^n, dx \
&le |K|lambda^n + varepsiloncdot1.
end{align}$$
Take limit as $ntoinfty$ yields
$$
limsup_{nto infty} int_{[0,1]} f(x)^n, dx le varepsilon.
$$
Since the above hold for any $varepsilon>0$, we have $int_{[0,1]} f(x)^n, dxto 0$ as wanted.