Constructing a continuous random variable on an atomless space

Question

Let $(\Omega,\mathcal F,P)$ be a GIVEN probability space which is atomless. Assume no topology or algebraic structure given on the space (but you can impose one on it) , how could I construct a (real-valued) continuous random variable on this space?

The only example I could think of are infinite sums of indicator functions of events, but that random variable would have countable image and hence discrete. It is simply challenging enough for me to construct a random variable with uncountable image. Anyone can help me? Thanks!

Edited: An atom in a space $(\Omega,\mathcal F,P)$ is a set $E\in \mathcal F$ such that $P(E)>0$ and that for each $F\subseteq E$, either $P(F)=0$ or $P(F)=P(E)$; A space is called atomless if it contains no atoms.

It is a fact that an atomless space must be uncountable, and by axiom of choice we can prove that for each $0\leq\alpha\leq 1$, there is some $E\in\mathcal F$ with $P(E)=\alpha$. How could I utilize this fact to construct such continuous random variable?

Answer 1

Note: Better answer below, leaving for posterity. Let's assume for now that $\Omega=\mathbb{R}$ and $\mathcal{F}$ is the Borel $\sigma$-algebra, and let $\mu$ be the given measure on $\mathbb{R}$. Then we can write $\mu((-\infty,x])=g(x)$ for some monotonic $g$, and as we have no atoms $g$ is continuous. Let $X^{-1}((-\infty,x])=(-\infty,f(x)]$ for some monotonic increasing $f(x)$ satisfying $f(\mathbb{R})=\mathbb{R}$, i.e. $\{X(\omega)\leq x\}\Leftrightarrow\{\omega\in(-\infty,f(x)]\}$. This is equivalent to defining $X(\omega)=y$ for all $\omega\in[\lim_{z\uparrow y}f(z),f(y)]$, which is well-defined if $f(\mathbb{R})=\mathbb{R}$. Then we can show that $\mathbb{P}[X\leq x]=g(f(x))$. So now the questions becomes, can we make $g(f(x))$ differentiable on all of $\mathbb{R}$ for arbitrarily weird functions $g$ that are continuous and satisfy $g(-\infty)=0$, $g(\infty)=1$?

I believe the answer is yes, and I've heard claims to that effect, but I can't find an actual proof. It's more of an analysis question, and an elementary enough one, so if you post a question with the analysis tag one of those guys might be able to help you out.

Update: Thanks to @Did for the clarification. It's sufficient to be able to construct a random variable uniformly distributed on the set $\{\frac{i}{2^n}\}_{i=1}^{2^n}$ for any $n$. Use the axiom of choice to inductively create the following partitions. For $n=1$, let $E_1^{(1)}$, $E_2^{(1)}$ partition $\Omega$ and satisfy $\mathbb{P}[E_i^{(1)}]=\frac{1}{2}$ for $i=1,2$. For general $n\geq 2$ and $1\leq k\leq 2^{n-1}$, let $E_{2k-1}^{(n)},E_{2k}^{(n)}$ partition $E_k^{(n-1)}$ into two sets of equal probability. Let $X_n(\omega)=\frac{1}{2^n}\sum_{i=1}^{2^n}i1_{\omega\in E_i^{(n)}}$. Then $X_n$ is uniformly distributed on the set $\{\frac{i}{2^n}\}_{i=1}^{2^n}$. Furthermore for every $\omega\in\Omega$ $X_n(\omega)$ is monotonically decreasing as $n\rightarrow\infty$, so $X_n$ converges pointwise (and therefore almost surely) to some random variable $X$. But almost sure convergence implies convergence in distribution so $X$ must be a $\text{Uniform}[0,1]$ random variable.