This answer gives more detail on my comments. In the following, I assume the axiom of choice whenever needed.
Claim 1 (counter-intuitive): There exists a probability space $(\Omega, \mathcal{F}, P)$ and two i.i.d. random variables $X:\Omega\rightarrow[0,1]$ and $Y:\Omega\rightarrow[0,1]$, both uniformly distributed on $[0,1]$, such that $X$ and $Y$ have disjoint images, that is $X(\Omega) \cap Y(\Omega) = \phi$.
Proof: Consider Theorem 3 from the following paper:
[D. Rizzolo, "Strange Uniform Random Variables," arxiv:1301.7148v1, 2013]
https://arxiv.org/abs/1301.7148
Theorem 3 there gives two probability spaces $(\Omega_1, \mathcal{F}_1, P_1)$ and $(\Omega_2, \mathcal{F}_2, P_2)$ with corresponding uniformly distributed random variables $Z_1:\Omega_1\rightarrow[0,1]$ and $Z_2:\Omega_2\rightarrow[0,1]$ (defined on their respective spaces) such that $Z_1(\Omega)\cap Z_2(\Omega)= \phi$. Define the product space $\Omega = \Omega_1\times \Omega_2$, $\mathcal{F} = \mathcal{F}_1\otimes\mathcal{F}_2$, $P=P_1\otimes P_2$ and define $X:\Omega\rightarrow[0,1]$ and $Y:\Omega\rightarrow[0,1]$ by
\begin{align}
X(\omega_1, \omega_2) &= Z_1(\omega_1) \quad \forall (\omega_1,\omega_2) \in \Omega\\
Y(\omega_1, \omega_2) &= Z_2(\omega_2) \quad \forall (\omega_1, \omega_2) \in \Omega
\end{align}
$\Box$
For the $X, Y$ random variables of Claim 1, it is impossible to have any function $f:\Omega\rightarrow\Omega$ (measurable or not) such that $X(\omega) = Y(f(\omega))$, because $X$ and $Y$ have disjoint images.
Claim 2: Suppose $(\Omega, \mathcal{F}, P)$ is a probability space and $X:\Omega\rightarrow[0,1]$ and $Y:\Omega\rightarrow[0,1]$ are random variables. Suppose both $X$ and $Y$ are uniformly distributed over $[0,1]$ and that both $X(\Omega)$ and $Y(\Omega)$ are Borel measurable subsets of $[0,1]$. Define $C=X(\Omega) \cap Y(\Omega)$. Then $C$ is Borel measurable, $P[X \in C] = 1$, and there is a (possibly nonmeasurable) function $f:\Omega\rightarrow\Omega$ such that $X(\omega) = Y(f(\omega))$ for all $\omega \in X^{-1}(C)$.
Proof: Since $X(\Omega)$ and $Y(\Omega)$ are Borel sets, the set $C$ is also Borel. Since $P[X \notin X(\Omega)] = 0$ and $P[Y\notin Y(\Omega)]=0$ we have
\begin{align}
P[X \notin C] &= P[\{X \notin X(\Omega)\}\cup \{X \notin Y(\Omega)\}]\\
&\overset{(a)}{\leq} P[X \notin X(\Omega)] + P[X \notin Y(\Omega)] \\
&\overset{(b)}{=}P[X \notin X(\Omega)] + P[Y \notin Y(\Omega)]\\
&=0
\end{align}
where (a) holds by the union bound; (b) holds because $Y$ has the same distribution as $X$. It follows that $P[X \in C] = 1$.
To finish the proof, since $P[X \in C] = 1$ we know $X^{-1}(C)$ is nonempty. Fix $\omega \in X^{-1}(C)$ and define $x=X(\omega)$. Then $x \in X(\Omega)\cap Y(\Omega)$. Since $x \in Y(\Omega)$, there is a $\nu \in \Omega$ such that $Y(\nu)=x$. Denote this $\nu$ by $\nu_{\omega}$, since it corresponds to our initial $\omega \in X^{-1}(C)$. So we have $x=X(\omega) = Y(\nu_{\omega})$. This holds for any arbitrarily chosen $\omega \in X^{-1}(C)$. So by the axiom of choice, for each $\omega \in X^{-1}(C)$ we have $\nu_{\omega} \in \Omega$ such that $X(\omega)=Y(\nu_{\omega})$. Define $f:\Omega\rightarrow\Omega$ by $f(\omega)=\nu_{\omega}$ for all $\omega \in X^{-1}(C)$ (and define $f(\omega)$ arbitrarily for $\omega \notin X^{-1}(C)$). $\Box$
Background: I found the Rizzolo paper last year when I had independently developed a related result on images of random variables (called "Claim 3" below). When I searched for similar prior work I found the Rizzolo paper and saw that his proof of his Prop 1 was essentially the same as my own proof! Of course, his proof was developed years earlier. The following Claim 3 should be viewed as a basic generalization of Claim 1 (and Claim 1 follows directly from Rizzolo).
Setup to Claim 3: Fix $F:\mathbb{R}\rightarrow\mathbb{R}$ as a valid cumulative distribution function (CDF), meaning it satisfies the basic properties of a CDF (nondecreasing, right-continuous, $\lim_{x \rightarrow\infty} F(x)=1$, $\lim_{x\rightarrow-\infty}F(x) = -1$). Define $\mathcal{B}(\mathbb{R})$ as the Borel sigma algebra on $\mathbb{R}$. Define $\mu_F:\mathcal{B}(\mathbb{R})\rightarrow[0,1]$ as the measure induced by CDF $F$. Specifically,
$$\mu_F(A)=P[X\in A] \quad \forall A \in \mathcal{B}(\mathbb{R})$$ whenever $X$ is a random variable (on some probability space) with CDF $F$. It is well known that any random variable with CDF $F$ produces the same measure, and so $\mu_F$ is well defined.
Definition: Given a CDF $F$, we say that a set $M\subseteq \mathbb{R}$ is an image of a random variable with CDF $F$ if there is a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X:\Omega\rightarrow\mathbb{R}$ with CDF $F$ such that $X(\Omega) = M$. Note that the set $M$ need not be Borel measurable.
Claim 3: Fix $F$ as a CDF. A set $M\subseteq \mathbb{R}$ is an image of a random variable with CDF $F$ if and only if $\mu_F(A)=0$ for all Borel measurable subsets $A \subseteq M^c$ (where $M^c = \mathbb{R}\setminus M$ is the complement of set $M$).
Proof: The forward direction is obvious: If $M$ is an image of some random variable $X$ with CDF $F$, so that $M=X(\Omega)$, then for any Borel measurable set $A \subseteq M^c$ we have $\{\omega \in \Omega : X(\omega) \in A\}=\phi$ so $P[X \in A]=0$. But since $X$ has CDF $F$ we also have $\mu_F(A)=P[X\in A]$. So $\mu_F(A)=0$.
For the reverse direction, suppose $F$ is a CDF and $M\subseteq\mathbb{R}$ is such that $\mu_F(A)=0$ whenever $A$ is a Borel measurable subset of $M^c$. Define $\Omega = M$ and $\mathcal{F}=\{M \cap A : A \in \mathcal{B}(\mathbb{R})\}$. It is not difficult to show $\mathcal{F}$ is indeed a sigma algebra on $M$. Now construct $P:\mathcal{F}\rightarrow[0,1]$ such that for each $D \in \mathcal{F}$, we can select any $A \in \mathcal{B}(\mathbb{R})$ for which $D=M \cap A$ and define $P[D]=\mu_F(A)$. The proof that this can be done consistently, and results in a valid measure, uses the axiom of choice and is almost identical to the proof of Prop. 1 in the Rizzolo paper. $\Box$
Claim 3 generalizes Claim 1 as follows: In the case when $F$ is the CDF of a random variable that is uniformly distributed over $[0,1]$, Claim 3 implies that if $M\subseteq[0,1]$ is a set such that $[0,1]\setminus M$ contains no Borel subset of positive measure, then $M$ is the image of some random variable that is uniformly distributed on $[0,1]$. Using Vitali set concepts, we can construct two disjoint sets $M_1\subseteq[0,1]$ and $M_2\subseteq[0,1]$ such that neither $[0,1]\setminus M_1$ nor $[0,1]\setminus M_2$ contains a Borel subset of positive measure. So Claim 3 implies existence of probability spaces $(\Omega_1, \mathcal{F}_1, P_1)$ and $(\Omega_2, \mathcal{F}_2, P_2)$ and uniformly distributed random variables $X_1:\Omega_1\rightarrow[0,1]$ and $X_2:\Omega_2\rightarrow[0,1]$ such that $X_1(\Omega) = M_1$, $X_2(\Omega)=M_2$. Since $M_1$ and $M_2$ are disjoint, $X_1$ and $X_2$ have disjoint images. Using the product space we obtain the result of Claim 1. $\Box$
