Why is $A(\mathbb{T})\subset C(\mathbb{T})$? where $A(\mathbb{T})$ is the space of Absolutely Converging Fourier Series and $C(\mathbb{T})$ is the space of Continuous Functions, both over $\mathbb{T} = [0,1)$. 
If $ f\in A(\mathbb{T})$ is $f\in C(\mathbb{T})?$
From the absolute convergence, one can prove the uniform convergence of the Fourier Series, thus the limit-function will be Continuous. But are we sure that the limit-function is f? We know that, if f is given Continuous. So I think that the result derives from something else, or I miss something about the Fourier Series and the corresponding function,
 A: Take an $f\in C^2(\mathbb{T})$. Then $f$ will obviously be continuous and its Fourier series will be absolutely convergent. Let 
$$ g(x) = \left\{
\begin{array}{ c l }
f(x),   &    x\in\mathbb{T}\setminus\{0\} \\
f(x)+1,   &    x=0
\end{array}
\right. $$
Then the Fourier series of $g$ will be the Fourier series of $f$ and thus will be absolutely convergent, but $g$ is not continuous. However, it is a.e. equal to a continuous function, which is the best you can achieve in general.
A: Here $s_n f(x)$ is the $n$-th partial Fourier series of $f(x)$.

Fejer's theorem
  Let $f \in \mathcal{L}_1(\mathbb{T})$.
  Then $\sigma_n f(x) \xrightarrow[n \rightarrow \infty]{} f(x)$ for all $x$ where $f$ is continuous. Moreover if $f \in \mathcal{C}(\mathbb{T})$ then $\sigma_n f \xrightarrow[n \rightarrow \infty]{\text{unif.}} f$ on $\mathbb{T}$.

Here $\sigma_n f(x) :=~ \frac{s_0 f(x) + ... + s_n f(x)}{n+1}$.

Corollary 1 (Uniqueness theorem for $\mathcal{C}(\mathbb{T})$ ) :  Let $f \in \mathcal{C}(\mathbb{T})$. If $\widehat{f}(k) = 0$ for all $k \in \mathbb{Z}$ then $f = 0$.
Corollary 2 : Let $f \in \mathcal{L}_1(\mathbb{T})$. If $s_n f(x) \xrightarrow[n \rightarrow \infty]{} L$ and if $f$ is continuous at $x$ then $L = f(x)$.

That is to say if the limit at $x$ exists and if $f$ is continuous at $x$ then the limit is "the right one".
