I have defined, for a probability measure $\eta$ we have the fourier transform as $\hat{\eta} = \int e^{itx} \ d\eta(x)$, and for a function $h: \mathbb{R} \to \mathbb{R}$ we have that the fourier transform of $h$ is given by $\hat{h}(t) = \int e^{itx} h(x) \ dx$.

I then am told that the fourier transform of a probability measure, is the fourier transform of its density if it has one.

I don't really understand this statement - what is the density of a measurable function? I thought random variables are those to have densities. Any explanation please, thanks.

  • $\begingroup$ The concept of measurable functions is much more general/generic than that of a probability density function. See this post, for example. This page discusses it a bit, and the common example of a random variable (or probability measure) without a density is the Cantor distribution. $\endgroup$
    – bgins
    Mar 28, 2016 at 21:32
  • $\begingroup$ I have just never seen what a density of a measure $\eta$ is, only a density of a random variable, which is a measureable function. $\endgroup$
    – lampj20la
    Mar 28, 2016 at 21:45

2 Answers 2


Maybe you are more familiar with the concept of Radon-Nikodym derivative. Basically the "density" is just the RN-derivative in the context of probability (see the "Application" section on the linked wiki page).

The claim is stating that if you can write $d\eta(x)$ as $f(x) dx$, where $dx$ is the Lebesgue measure, then

$$ \hat \eta (t) = \int e^{itx} d\eta(x) = \int e^{itx} f(x) dx = \hat f(t) $$


A random variable need not be continuous. A random variable is a measurable function $$X:(\Omega,\mathscr{A})\to (\mathbb{R},\mathscr{B}).$$

A measurable function "transports" a measure from $(\Omega,\mathscr{A})$ to $(\mathbb{R},\mathscr{B})$ as follow $$\mathbb{P}_X(B)=\mathbb{P}(X\in B),\quad A\in\mathscr{B}.$$

Now for a measurable function $u:\mathbb{R}\to\mathbb{R}$, you can integrate the composition $u\circ X$ with respect to this image measure as follows $$\int_{\Omega} u(X(\omega))\mathbb{P}(d\omega)=\int_{\mathbb{R}}u(x)\mathbb{P}_X(dx).$$

Now a random variable is absolutely continuous, if its distribution, this image measure $\mathbb{P}_X$ is (absolutely) continuous with respect to the Lebesgue measure, meaning there exists a positive measurable, $\mathbb{P}_X$ almost everywhere uniqure $f$, such that: $$\mathbb{P}_X(A)=\int_A f(x)\lambda(dx).$$ This $f$ is the density of $X$ or the Radon-Nikodym derivative of $\mathbb{P}_X$ with respect to $\lambda$. Hence its characteristic function would be: $$\phi_X(\xi):\mathbb{E}e^{i\xi X}=\int_\Omega e^{i\xi X(\omega)}\mathbb{P}(d\omega)=\int_\mathbb{R} e^{i\xi x}\mathbb{P}_X(dx)=\int_\mathbb{R}e^{i\xi x}f(x)\lambda(dx).$$


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