Probability measures are representable through measurable functions?

Question

Let $f$ be a measurable function $f:D\to \mathbb C$, where $D$ is a measurable set in $\mathbb R^k$ with finite non-zero measure. We know that the linear positive operator $\Lambda:C_c(\mathbb C) \to \mathbb R$ given by $$ \Lambda(G) = \frac{1}{|D|}\int_D G(f(x)) dx $$ is (uniquely) representable by a radon probability measure $\mu$ such that $$ \Lambda(G) = \int_{\mathbb C} G d\mu $$ I wonder if the opposite is also true. That is

Given a probability measure $\mu$ on $\mathbb C$, there exists a measurable function $f:[0,1]\to \mathbb C$ s.t. $$ \int_{\mathbb C} G d\mu = \int_0^1 G(f(x)) dx$$ for every $G\in C_c(\mathbb C)$?

It's easy to build $f$ if $\mu$ is atomic, but I don't know if it is true for singular or continuous measures.

Answer 1

I would check this page: https://en.wikipedia.org/wiki/Standard_probability_space ... See the first paragraph of the section verifying the standardness, which says that any probability measure on euclidean space is standard.

This means that any probability measure on euclidean space is isomorphic (up to null sets) to a disjoint union of an interval and a countable atomic measure. (Isomorphism here means in the category of measure spaces with measurable measure preserving maps. Up to null sets means that we are allowed to construct this isomorphism on the complement of sets of measure zero... this is written better on wikipedia.)

I think I can prove this but my argument below reflects the messy state of my mind right now. I'll try to clean it up and verify details after I get some sleep...

A strategy for proof would be to reduce it to the case of an interval (is it easy to show that the class of standard probability spaces is closed under products?* this is stated as true on wikipedia), and then use the cumulative density function to prove this in the case of the interval. (I see now that kimchilover is discussing this. The case of the interval is a pretty standard fact from measure theoretic probability courses, it should be done in detail in Durret, if I recall correctly.)

(*Ahout the case of products - applying the guaranteed decomposition into interval and atomic part to the factors and using distributivity of the product, this reduces to showing that the Lebesgue measure on $I^n$ is a standard probability space. This would follow from induction after proving the existence of a measure preserving map $I \to I^2$. I think you can produce one as a the limit of a sequence of maps $f_n : I \to I^2$, where $f_n$ breaks $I$ into $2^n$ pieces and puts them in $I^2$ in a checkerboard fashion, and $f_{n+1}$ refines $f_n$ so that there is a well defined pointwise limit.)

Given that, I think a yes to your question is immediate - you replace the complex numbers with the interval $J$ and some atoms (after removing a null set), chop up your parametrizing interval $[0,1]$ into two appropriately sized intervals $I_1$ and $I_2$, and use $I_2$ to produce the atoms (again by chopping it up into intervals), and parametrize $J$ with $I_1$.

Answer 2

Yes. You are basically asking, given a probability measure $\gamma$ (I'm changing notation on you here, a bit) on $\mathbb C$, is there a random variable with $\gamma$ as its probability distribution?

Recall the trick that if a real random variable has distribution function $F$, it can be simulated by $F^{-1}(U)$, where $U$ is uniformly distributed on $[0,1]$ as follows, where an obvious convention about the meaning of $F^{-1}$ handles jump discontinuities in $F$. You are in effect asking for a 2 dimensional version of this.

There might be elegant ways to show this, but this is what comes to the top of my head. Let $\pi$ be the marginal distribution of $\operatorname{Re}(z)$ when $z\sim \gamma$. That is, it is the image of $\gamma$ induced by $z\mapsto \operatorname{Re}(z).$ Disintegrate $\gamma$ with respect to $\pi$, writing $\gamma=\int_{\mathbb R} \gamma_x \pi(dx)$ where $\gamma_x$ is the conditional distribution of $\operatorname{Im}(z)$ given $\operatorname{Re}(z)=x.$ (See the wikipedia article about this.)

Let $(U,V)$ be uniformly distributed on $[0,1]\times[0,1]$, and let $X = F^{-1}(U)$ where $F$ is the distribution function for the measure $\pi$ and $Y = G_X^{-1}(V)$, where $G_x (y)=\gamma_x((-\infty,y])$ is the distribution function of $\gamma_x$. Write the composition as $\phi:(U,V)\mapsto (X,Y)$. Finally, let $f=\phi\circ m$ where $m$ is any Lebesgue-preserving map from $[0,1]\to[0,1]\times[0,1]$. Then the image of Lebesgue measure on $[0,1]$ under $f$ is your desired $\gamma$.