Is $A(\Bbb T) \subset C(\Bbb T)$?

Question

My instructor defined the space $A(\Bbb T)$ as $$A(\Bbb T) := \left\{f\in L^1(\Bbb T): \sum_{n=-\infty}^\infty |\hat f(n)| < \infty\right\}$$ and wrote $A(\Bbb T) \subset C(\Bbb T)$, where $C(\Bbb T)$ denotes the space of continuous functions on $\Bbb T = \Bbb R/\Bbb Z$. I'm trying to make sense of this claim - even though it doesn't seem true in the first place - as two functions $f,g$ that are equal a.e. have the same Fourier coefficients. If $f$ is a continuous function in $A(\Bbb T)$, we can find another function $g \ne f$, such that $g = f$ a.e., and $g$ is not continuous (for example, change the value of $f$ at one point.) As $g = f$ a.e., $$\sum_{n=-\infty}^\infty |\hat g(n)| = \sum_{n=-\infty}^\infty |\hat f(n)| < \infty.$$

So, what's the instructor trying to say? Is it that for every function $f \in A(\Bbb T)$, there is a continuous function $g$ such that $g = f$ a.e.?

Consider $g(t) := \sum_{n=-\infty}^\infty \hat f(n) e^{2\pi int}$ where $\sum_{n=-\infty}^\infty |\hat f(n)| < \infty$, i.e., $f \in A(\Bbb T)$. Then, $g$ is continuous, as it is a uniform limit of continuous functions (its partial sums converge uniformly to $g$, by the Weierstrass M-Test.) Furthermore, $\hat g(n) = \hat f(n)$ for all $n \in \Bbb Z$. This gives $f = g$ a.e.

By the way, Katznelson's Harmonic Analysis defines $A(\Bbb T)$ as a subset of $C(\Bbb T)$ instead of $L^1(\Bbb T)$, in Section $1.6.1$:

Thanks for clarifying!

Answer 1

As @user10354138 says, this is a matter of convention.

In order to understand the construction of $L^1$, which is important, we need to identify all functions that are equal almost everywhere. However,

• To always talk about the elements of $L^1$ as equivalence classes would be clumsy.
• We learned about functions long before we learned about Lebesgue integration and we don’t want leave the idea of functions.
• The reason to introduce Lebesgue integration was not superficious or for fun, it was introduced because it was needed in order to understand limits of functions.

Moreover, I myself look at $A(\mathbb{T})$ both as an algebra of nice functions, and as the algebra $\ell^1(\mathbb{Z})$ of sequences. Probably in the same manner as you look at $\mathbb{T}$ as the unit circle, the quotient space $\mathbb{R}/\mathbb{Z}$ or the intervall $(-\pi,\pi)$.