Property of inverse of Fourier transform

Question

I have studied the Fourier transform and the inverse Fourier transform for functions in $L^1(\mathbb{R})$, so in 1D. In my book, I see the following definition:

If $f \in L^1(\mathbb{R})$ and its Fourier transform $\hat{f} \in L^1(\mathbb{R})$, then the inverse Fourier transform of $\hat{f}(\omega)$ is defined by: $g(t) = \mathcal{F}^{-1}\{\hat{f}(\omega)\} = \dots $.

Additionally, $g=f$ for almost every $t \in \mathbb{R}$. If the original $f$ is continuous, then $g=f$ for every $t \in \mathbb{R}$.

I don't understand how that is possible. I imagine that a function in $f \in L^1(\mathbb{R})$ could be discontinuous almost everywhere. But by applying Fourier transform and inverse Fourier transform, I will receive a function $g$ that is continuous on $\mathbb{R}$.

By the proposition above, this would be a fatal issue.

What am I thinking wrong?

Answer 1

You are absolutely right that there are functions $f\in L^1$ sucht that there is no continuous function $g$ with $f=g$ almost everywhere.

Nevertheless, the statement that you cite is true: If in addition to $f\in L^1$ you assume $\hat{f}\in L^1$, then $f=\mathcal{F}^{-1}\hat{f}$ almost everywhere, where the right hand side is continuous.

In particular, this shows that if $f$ is such that there is no continuous $g$ with $f=g$ almost everywhere, then you must have $\hat{f}\notin L^1$.

Answer 2

The Fourier inversion theorem holds for all Schwartz functions (roughly speaking, smooth functions that decay quickly and whose derivatives all decay quickly). I think this is one of the regularity conditions f can obey.

Answer 3

The proper statement of this is that if $f\in L^1$ and $\hat f \in L^1$, then $\mathcal F^{-1}(\hat f)(x) = f(x)$ for almost every $x$. Since the inverse Fourier transform of an $L^1$ function is continuous, $f$ can be modified on a set of measure zero so that it is continuous, but it need not be continuous, as you already noted.

If the original function $f$ is also continuous, then we in fact gain that $\mathcal F^{-1}(\hat f)(x) = f(x)$ for every $x$ because two continuous functions that agree almost everywhere must agree everywhere.