What does it mean by "Brownian motion is not the distribution function of any signed measure"?

Question

Could someone explain me what does it mean by "Brownian motion is not the distribution function of any signed measure"? I was reading some notes on Stochastic calculus where I encountered this statement which is not vert clear to me,

Answer 1

Any distribution function is almost everywhere differentiable, because it is monotone. Brownian motion is nowhere differentiable, so it cannot be a distribution function of any signed-measure.

Answer 2

A real function $W$ is the distribution of a signed measure $m$ if the measure of $(a,b]$ equals $W(b)-W(a)$ for every real numbers $a<b$.

If $W$ is random and given by a Brownian motion, it is a continuous function and you could wonder whether this random function defines a random signed measure $m$. The statement quoted in the question is saying it (almost surely) does not.

One way to show that there is (a.s.) no signed measure whose distribution is given by $W$ is the following. Distribution functions of signed measures must have bounded variation, and the Brownian motion a.s. does not.