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$$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$ when f is continuous on $[0,1]$

I know it can be proved using bounded convergence theorem but, I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ... Thank you.

quicksilver
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3 Answers3

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take any $\epsilon$, choose $\delta > 0$ such that $|f(x) - f(0)| < \epsilon$ on $[0,\delta]$. choose $n$ big enough such that $(1-\epsilon)^n < \delta$ then $$ |\int_0^1 f(x^n) dx - f(0) |= |\int_0^{(1-\epsilon)} [f(x^n) - f(0)]|dx + \int_{1-\epsilon}^1 [f(x^n) - f(0) ]dx | \leq $$ $$ \int_0^{(1-\epsilon)} |f(x^n) - f(0)|dx + \int_{1-\epsilon}^1 |f(x^n) - f(0) |dx $$ first factor is smaller than $\epsilon(1 - \epsilon)$ thanks to the choice of $\delta$ and $n$, second one is smaller than $\epsilon \cdot 2 \sup |f|$ because length of your interval of integration is $\epsilon$ so the result follows since $\epsilon$ was arbitrarily small

mm-aops
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Hint:

$f(0)=\int_0^1 f(0) dx$

$x^n\to 0$ for $0\leq x<1$

and $f$ is continuous.

Haha
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    Technically you need that $f$ is uniformly continuous to pull the limit inside the integral (which is itself a limit). In this case, that is guaranteed by continuity of $f$ on $[0,1]$ and compactness of $[0,1]$. A detail which might be okay to omit for a student far enough along to know the dominated convergence theorem, but I feel it is worth noting it for those who do not. – zibadawa timmy Nov 18 '13 at 21:39
  • @zibadawatimmy that's what i meant:) – Haha Nov 19 '13 at 01:12
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Handsome tip:

Prove it for $g\left(x\right)=f\left(x\right)-f\left(0\right)$ where $g\left(0\right)=0$ and then make use of $\int_{0}^{1}f\left(x^{n}\right)dx=\int_{0}^{1}g\left(x^{n}\right)+f\left(0\right)dx=\int_{0}^{1}g\left(x^{n}\right)dx+f\left(0\right)$.

drhab
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