We have $\hat f(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{it})e^{-int}dt$ for each $f\in L^2(T)$ and for each $n\in \mathbb Z$.
Now define $\Lambda_n f=\sum_{k=-n}^{n}\hat f(k)$ and set $S=\left \{ f:\lim \Lambda_n f \ \text {exists}\right \}$.
The claim is that $S$ is dense in $L^2(T)$ and of the first category.
Here is my argument:
For the first part, considering real and imaginary parts separately, using the definition of the Lebesgue measure and a density argument, it suffices to take $f=\chi_{(a,b)}$, the characteristic function of the interval $(a,b)\subset (-\pi, \pi).$. Then, $\int_{-\pi}^{\pi}\chi_{(a,b)}\cos {nt}dt=\frac{1}{n}(\sin nb-\sin na)$ and similarly for $\int_{-\pi}^{\pi}\chi_{(a,b)}\sin {nt}dt$.
By Dirichlet's Test, using the fact that $\sum_{n=1}^m \sin(nt)=\frac{\sin(mt/2)\sin((m+1)t/2)}{\sin (t/2)},\ $ so that $\vert\sum_{n=1}^m \sin(nt)\vert\leq \frac{1}{\vert\sin (t/2)\vert},\ $it's not hard to show that these series converge. This proves that $\chi _{(a,b)}\in S.$
For the second part, we observe that if $S$ is of the second category, then we may apply Banach-Steinhaus to $\left \{ \Lambda_n \right \}_n$, to say that $S=L^2(T)$, which is a contradiction because it is well-known that there are Fourier series that are not convergent in $L^1$; i.e. the Fourier coefficients are not summable.
My questions are: is my proof correct? If not, can you supply a hint or two? Is there a more elementary way to prove the claim?
edit: the exercise as posed, does not assume any knowledge of Fourier Analysis.