Integrable Fourier transform implies continuity I have read the following statement in a number of places  (here for example)

A discontinuous function cannot have an integrable Fourier transform.

The contrapositive of this statement is: if $\hat f$ is in $L^1(\mathbb R)$, then $f$ is continuous. I fail to see why this is true. I know that if $\hat f$ is in $L^1(\mathbb R)$, then $f=\mathcal F^{-1}(\hat f)$ a.e. The inverse Fourier transform $\mathcal F^{-1}(\hat f)$ is indeed a continous function, but how can continuity of $f$ be derived ? A discontinuous function may coincide with a continuous function except at a single point...
In a more general setting:
 A: I'll abuse notation and write $u(x) \in L^1(\mathbb R)$ for an expression $u(x)$ instead of writing the more correct $x \mapsto u(x) \in L^1(\mathbb R)$
Suppose that $\hat f(\xi) \in L^1(\mathbb R)$. Then also $\hat f(\xi) e^{i \xi x} \in L^1(\mathbb R)$.
By the Fourier inversion formula,
$$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi$$
We want to show that for any $x_0 \in \mathbb R$ we have $\lim_{x \to x_0} f(x) = f(x_0)$.
It's enough to show that $\lim_{x_n \to x_0} f(x_n) = f(x_0)$ for any sequence $x_n \to x_0$.
Let $\hat f_n(\xi) = \hat f(\xi) e^{i \xi x_n}$ and $g(\xi) = |\hat f(\xi)|$. Since $e^{i \xi x}$ is continuous in $x$, we have pointwise convergence, i.e. $\hat f_n(\xi) \to \hat f(\xi)$ for every $\xi \in \mathbb R$. Also, for all $n$, $|\hat f_n(\xi)| \leq g(\xi)$ where $g(\xi) = |\hat f(\xi)| \in L^1(\mathbb R)$.
By the dominated convergence theorem then $\int \lim_{n \to \infty} \hat f_n(\xi) d\xi = \lim_{n \to \infty} \int \hat f_n(\xi) d\xi$, i.e.
$$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi 
= \frac{1}{2\pi} \lim_{n \to \infty} \int \hat f_n(\xi) d\xi
= \lim_{n \to \infty} \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x_n} d\xi
= \lim_{n \to \infty} f(x_n)$$
Thus, $f$ is continuous.
