Let $f:[0,1] \rightarrow \mathbb R$ be Riemann integrable and continuous at 0. Show that $\lim_{n \rightarrow \infty}\int_{0}^{1}f(x^{n})dx=f(0).$
My attempt: Put $x^{n}=u$, $nx^{n-1}dx=du$ and $dx=\frac{1}{n}u^{\frac{1}{n}-1}du$.
Now, $|\int_{0}^{1} \frac{1}{n} f(u) u^{\frac{1}{n}-1}du-f(0)| \leq\int_{0}^{1} |\frac{1}{n} (f(u)-f(0)) u^{\frac{1}{n}-1}du|$.
Since $f$ is continuous at 0. So for given $\epsilon>0$, choose a $\delta$ such that $|u-0|<\delta$. Therefore,$|f(u)-f(0)|<\epsilon$.
The above equation implies that $|\int_{0}^{1} \frac{1}{n} f(u) u^{\frac{1}{n}-1}du-f(0)| \leq \epsilon [u^{\frac{1}{n}}]_{0}^{1}$.
Hence prove the given result.
Is there is any mistake in this proof or any other justification be required?
Since $f$ is Riemann integrable and ,hence, bounded, there exists $M$ such that $|f(x)| \leqslant M$ for all $x \in [0,1]$.
Note that
$$0 \leqslant\left|\int_0^{1} f(x^n) \, dx - f(0)\right| \leqslant \int_0^{\delta^{1/n}} |f(x^n) - f(0)|\, dx + \int_{\delta^{1/n}}^1|f(x) - f(0)| \, dx \\ \leqslant \int_0^{\delta^{1/n}} |f(x^n) - f(0)|\, dx + \int_{\delta^{1/n}}^1(|f(x)| +|f(0)|) \, dx\\ \leqslant \int_0^{\delta^{1/n}} |f(x^n) - f(0)|\, dx + 2M(1- \delta^{1/n})$$
You should be able to finish from here using what you know about the continuity of $f$ at $x=0$.