fn, are continuous on [0,1] and converges pointwisely to f,
and,
$\displaystyle \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x) }$ again,
must f be continuous ?
give counterexample
Here is an easier conterexample.
Let $\displaystyle f_n(x) = x^n$ then each one is continous and converges to $\displaystyle f(x)$ which is $\displaystyle 0$ for $\displaystyle 0\leq x < 1$ with $\displaystyle f(1) = 1$.
Now $\displaystyle \int_0^1 x^n dx = \frac{1}{n+1} \to 0 = \int_0^1 f(x) dx$
But $\displaystyle f$ is not continous.
Reduction formula: $\displaystyle \int \sin^n(\pi x)\,dx = -\frac{1}{n}\cos(\pi x)sin^{n-1}(\pi x) + \frac{n-1}{n}\int sin^{n-2}(\pi x)\,dx$
There are some $\displaystyle \pi$'s and $\displaystyle \frac{1}{\pi}$'s that should be in there somewhere, but I'm too lazy to find out where, and they shouldn't make any difference anyway. As $\displaystyle n\to\infty$, the non-integral part clearly goes to 0, and then you might be able to prove by induction that everything is 0. Or something.
If $\displaystyle t=\pi x$ then we get $\displaystyle \frac{1}{\pi}\int_0^{\pi} \sin^n t dt$.
The function $\displaystyle \sin^n t$ is symmetric about $\displaystyle t=\frac{\pi}{2}$.
Thus, $\displaystyle \frac1{\pi}\int_0^{\pi/2}\sin^n t = \frac{2}{\pi}\int_0^{\pi/2}\sin^n t dt = {{2n}\choose n}\cdot \frac{1}{2^{2n}}$
It remains to show that expression goes to zero (such as using Stirling's approximation).
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Notice something interesting, there is a power series:
$\displaystyle \sum_{n=0}^{\infty} {{2n}\choose n}z^n = \frac{1}{\sqrt{1-4z}}$ for $\displaystyle |z|<\tfrac{1}{4}$
Therefore, $\displaystyle {{2n}\choose n}z^n \to 0$ if $\displaystyle |z| < \tfrac{1}{4}$.
But in our case $\displaystyle z=\tfrac{1}{4}$ to get $\displaystyle {{2n}\choose n}\cdot \frac{1}{2^{2n}}$.
However, $\displaystyle \sum_{n=0}^{\infty} {{2n}\choose n}z^n$ cannot be analytically continued at $\displaystyle z=\tfrac{1}{4}$ since $\displaystyle \frac1{\sqrt{1-4z}}$ explodes that at point.
Somethings seems to be wrong.