I'm trying to understand this problem: Let $f$ be Riemann integrable on $[0,2\pi]$
Suppose that the Fourier Series of $f$, $S_{n}^{f}(x)$, converges uniformly on the interval.
I want to show that $S_{n}^{f}(x)\to f(x)$ point-wise almost everywhere.
My attempt:
Since $S_{n}^{f}(x)$ converges uniformly, and $S_{n}^{f}(x)$ is continuous then it holds that $S_{n}^{f}(x) \to g(x)$ uniformly for $g$ continuous.
Then clearly $S_{n}^{f}(x_{0})$ converges to $g(x_{0})$ everywhere on the interval.
Also, since by definition $\int (S_{n}^{f}(x)-f(x)^2dx \to 0$
then $\int (lim_{n\to \infty}S_{n}^{f}(x)-f(x)^2dx = 0 = \int (g(x)-f(x)^2dx$
thus $g=f$ almost everywhere.
Is this reasoning correct?
thanks!