What does it mean u(dx) in the Fourier transform of a probability measure u?

Question

Let $\mu$ be a probability on $\mathbb{R}^n$ and consider its Fourier transform

$\overset{\wedge}{\mu} (u) = \int e^{i (u ,x)} \mu( dx)$,

where $(u, x)$ is the scalar product of $u$ and $x$.

What does it mean $\mu( dx)$?

Thank you!

Answer 1

It is just notation.

Think of it as the measure of the 'infinitesimal' slice $dx$. I like the notation in that it is suggestive of a Darboux sum, but it is a little cumbersome.

Alternative notations are $\int e^{i(u,x)} d \mu(x)$, or $\int f d\mu$ with $f(x) = e^{i(u,x)}$.

Answer 2

It is more intuitive to think $\mu(dx)$ in the form of $p(x) dx$.

$\mu$ is a probability measure and generally it is assumed to be absolutely continuous with respect to Lebesgue measure $dx$. So let's put (informally) $\mu(dx) = p(x) dx$ where $p(x)$ is the probability density function of this probability measure and $dx$ is the Lebesgue measure.

Then $\hat{\mu}$ can be written in the following form: $\hat{\mu} = \mathbb{E}[e^{i(u,x)}]$.