Yes, trivially: consider the probability space $\Omega = \mathbb{R}$, $\mathcal{F} = \mathcal{B}$, $\mathbb{P} = \mu$, and let $X : \Omega = \mathbb{R} \to \mathbb{R}$ be the identity map $X(\omega) = \omega$.
Alternatively, use inverse transform sampling. Let $F(x) = \mu((-\infty, x])$ be the cumulative distribution function of $\mu$, and let $G(t) = \inf\{x : F(x) \ge t\}$ be its "inverse". (If $F$ is actually 1-1 then $G$ is truly the inverse of $F$.) Now let $U$ be a Uniform(0,1) random variable on any probability space $(\Omega, \mathcal{F}, \mathbb{P})$ (e.g. the identity map on $[0,1]$ with Lebesgue measure) and set $X = G(U)$. It is then easy to see that the measure induced by $X$ is again $\mu$. Proof: we have $\mathbb{P}(X \le x) = \mathbb{P}(G(U) \le x) = \mathbb{P}(U \le F(x)) = F(x)$. So the measure induced by $X$ has $F$ as its cdf, and this uniquely determines it as being $\mu$.
There was such a question: "Is there anyway that these two types of convergence are related?"
Suppose we have $X_n \to X$ almost surely.
Put $P_n (A) = P(X_n \in A)$ and $P(A) = P(X \in A)$ for all Borel set $A$. Then $P_n$, $n \ge 1$ and $P$ are probability measures on real line.
$X_n$ converges to $X$ almost surely, hence $X_n$ converges to $X$ in distribution, and a sequence of measures $P_n$ converges weakly to the measure $P$. According to Portmanteau Theorem (see Billingsley, Convergence of probability measures, 1999, 2nd edition, pp.15-16) it follows that $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$. The next example shows that condition $P_n(A) \to P(A)$ holds not for all Borel sets $A$.
Consider a probability space $(\Omega, \mathcal{F}, \mu)$ where $\Omega = [0,1]$, $\mathcal{F}$ is sigma-algebra of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$. Put $\xi_n(\omega) = n \omega $ for $\omega \in [0, \frac{1}n ]$ and $\xi_n(\omega) = 0$ otherwise. In this case $P_n \sim U[0, \frac{1}{n}]$, $\xi_n \to X \equiv 0$, $P$ is Dirac measure at zero.
For $A = [ - 1, 0] $ we have $P_n(A) = 0 \nrightarrow P(A) = 1$.
As a result we get the next statement. If $X_n$ converges to $X$ almost surely then for corresponding measures $P_n$ and $P$ we have $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$ but not for all $A$.
Now suppose we have convergence $P_n(A) \to P(A)$ for all $A$ from some sigma-algebra. If these measures are not probability measures of if these measures are not measures on $\mathbb{R}$ with Borel sigma-algebra, then there is no direct relationship between them and any random variables. Suppose that $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel sets $A$. Hence $P_n$ converges to $P$ weakly (Portmanteau Theorem). According to the Skorokhod's representation theorem we can construct random variables $X_n$, $X$ such that $X_n$ has distribution $P_n$, $X$ has distribution $P$, $X_n \to X$ almost surely and moreover $X_n$ and $X$ are defined of probability space $(\Omega, \mathcal{F}, \mu)$, where $\Omega = [0,1]$, $\mathcal{F}$ consists of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$.
As a result we get the next statement. If $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel set $A$ then there are random variables $X_n, X$ on a "simple" probability space such that $X_n$ has distribution $P_n$, $X$ has distribution $P$ and $X_n \to X$ almost surely.
Best Answer
Suppose to the contrary that the statement does not hold. Then $$\Bbb{P}(\xi< \eta) = \Bbb{P}(\xi > \eta) = 0$$ which means that $\Bbb{P}(\xi = \eta) = 1$. Thus $\xi = \eta$ almost surely. But since $\xi$ and $\eta$ are independent, we have $$\Bbb{P}(\xi \in A, \eta\in B) = \Bbb{P}(\xi \in A) \Bbb{P}(\eta \in B) $$ for Borel sets $A,B$. Taking $A=B$, and using that $\xi =\eta$ almost surely we obtain $$\Bbb{P}(\xi \in A) = \Bbb{P}(\xi \in A)^2$$ such that $\Bbb{P}(\xi \in A)\in \{0,1\}$ for every Borel set $A$. Such a random variable must be constant almost surely, contradicting non-degeneracy.