How to reconcile the Fourier inversion on probability measures and on tempered distributions?

In probability theory, the Fourier Inversion theorem for a Borel probability measure $$\mu$$ on $$\mathbb{R}$$ reads:

$$\mu((a,b)) + \frac{1}{2} \mu(\{ a,b \}) = \frac{1}{2\pi} \lim_{T\to\infty} \int^T_{-T}\int^b_a \hat{\mu}(t) e^{-ixt} dx dt$$

On the other hand, in distribution theory, the Fourier Inversion theorem for a tempered distribution $$u$$ on $$\mathbb{R}$$ is exceptionally simple:

$$u = \frac{1}{2\pi} \mathcal{R}(\hat{\hat{u}})$$

where $$\mathcal{R}$$ is the reflection operator. Of course, this identity has to be understood in the distributional sense.

My question is: how to reconcile these two inversion theorems?

My try:

First, note that any Borel probability measure $$\mu$$ is a tempered distribution with Fourier transform given by:

$$\hat{\mu}(t) = \int_\mathbb{R} e^{ixt} d\mu(x)$$

By the Fourier Inversion theorem for a tempered distribution, for any Schwartz function $$\psi \in \mathcal{S}(\mathbb{R})$$:

\begin{align*} \int_\mathbb{R} \psi d\mu &= \frac{1}{2\pi} \langle \mathcal{R}(\hat{\hat{\mu}}), \psi \rangle = \frac{1}{2\pi} \langle \hat{\mu}, \widehat{\mathcal{R}\psi} \rangle \\ &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\hat{\psi})(t) dt \\ &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \int_\mathbb{R} \psi(x) e^{-ixt} dx dt \end{align*}

Next I want to "substitute" $$\psi = \chi_{(a, b)}$$, but here are two problems:

1. The function $$\chi_{(a, b)}$$ is not a Schwartz function. I guess this is where we cook up the limit $$T \to \infty$$.
2. I don't know how the term $$\frac{1}{2} \mu(\{ a,b \})$$ pops up.

Remark: To be consistent with the notation in probability theory, I define the Fourier transform of an integrable function $$f$$ on $$\mathbb{R}$$ to be

$$\hat{f}(t) = \int_{-\infty}^\infty f(x) e^{ixt} dx$$

• Instead we can go the other way and prove the Fourier inversion theorem (for L^1 functions, maybe it’s more subtle for distributions) from the probabilistic version. Commented Apr 10, 2023 at 14:52
• Have you seen a proof of the probabilistic inversion formula? (one that doesn’t use Fourier analysis?) Because it’s not so bad and the “$1/2$” term comes out nicely from the proof that I know. I sketch the proof in my linked answer and can reference a full proof or write one up if you wish Commented Apr 10, 2023 at 14:53
• @FShrike Yes, I have read the proof in your linked answer. It seems to prove by directly computing the double integral in the inversion formula. Unfortunately I can't see how such direct computation is related to the distributional Fourier inversion. Commented Apr 10, 2023 at 15:37

I am the questioner. I think I might have an answer.

Let's pretend we don't know the identity $$u = \frac{1}{2\pi} \mathcal{R}(\hat{\hat{u}})$$ and try to compute directly. For any Schwartz function $$\psi \in \mathcal{S}(\mathbb{R})$$, we have

\begin{align*} \frac{1}{2\pi} \langle \mathcal{R}(\hat{\hat{u}}), \psi \rangle &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\hat{\psi})(t) dt \\ &= \frac{1}{2\pi} \int_\mathbb{R} \int_\mathbb{R} e^{ixt} d\mu(x) \mathcal{R}(\hat{\psi})(t) dt \\ &= \int_\mathbb{R} \frac{1}{2\pi} \int_\mathbb{R} e^{ixt} \mathcal{R}(\hat{\psi})(t) dt d\mu(x) \end{align*}

The last step is valid for $$\psi \in \mathcal{S}(\mathbb{R})$$ because $$\int \int |e^{ixt} \mathcal{R}(\hat{\psi})(t)| dt d\mu = ||\mu|| ||\hat{\psi}||_1 < \infty$$

Now I want to substitute $$\psi = \chi_{(a, b)}$$, the problem is that $$\chi_{(a, b)}$$ is not a Schwartz function and $$\widehat{\chi_{(a, b)}}$$ is not integrable. This is where we need the limit $$T \to \infty$$ to permit the interchange of the order of integration.

\begin{align*} \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt &= \frac{1}{2\pi} \lim_{T \to \infty} \int_{-T}^T \int_\mathbb{R} e^{ixt} d\mu(x) \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt \\ &= \lim_{T \to \infty} \int_\mathbb{R} \frac{1}{2\pi} \int_{-T}^T e^{ixt} \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt d\mu(x) \end{align*}

Since $$\chi_{(a, b)}$$ is piecewise smooth, classical Fourier analysis tells us that we have the pointwise convergence for each $$x \in \mathbb{R}$$:

$$\Psi_T(x) := \frac{1}{2\pi} \int_{-T}^T e^{ixt} \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt \\ \to \frac{\chi_{(a, b)}(x+) + \chi_{(a, b)}(x-)}{2} = \chi_{(a, b)} + \frac{\chi_{\{a, b\}}}{2}$$

By a routine calculation, one can express $$\Psi_T$$ in terms of sinc functions and show that it is uniformly bounded. Thus, by dominated convergence, we have

$$\lim_{T \to \infty} \int_\mathbb{R} \Psi_T d\mu =\int_\mathbb{R} \chi_{(a, b)} + \frac{\chi_{\{a, b\}}}{2} d\mu$$

The result follows.