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

$$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$ 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.

Solution 1:

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

Solution 2:

Hint:

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

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

and $f$ is continuous.