I will try to write my question with every possible detail because I get very confused with these things, so I apologise for being too wordy.
Consider the space $\Omega := C([0, T], \mathbb{R})$ of contiuous functions from $[0, T]$ to the line. I will endow $\Omega$ with the sigma algebra $\mathcal{F} = \sigma(X_t)_{t \in [0, T]}$, where $$ X_t: \Omega \to \mathbb{R}, \quad \omega \to \omega(t). $$ That is, $\mathcal{F}$ will be the smallest $\sigma$-algebra on $\Omega$ which makes every evaluation function $X_t$ measurable. Now, for the very special sets $B \in \mathcal{F}$ of the form $B = X_t^{-1}(a, \infty)$, I will define
$$ \mathbb{P}(B) := \tilde{\mathbb{P}}(\mathcal{N}(0, t) > a) = \frac{1}{\sqrt{2\pi t^2}}\int_a^{\infty}e^{\frac{-x}{2t^2}}dx, $$ where $\tilde{\mathbb{P}}$ is the probability of a normal random variable andand $\mathcal{N}(0, t)$ is the nomal random variable of mean zero and variation $T$. Use these sets $B$ to extend $\mathbb{P}$ to a probability measure in the whole measure space $(\Omega, \mathcal{F})$ (I'm not completely sure if this is possible due to technical details, but that is not the point here). In brief, I understand the space $(\Omega, \mathcal{F}, \mathbb{P})$ as being the canonical space for the definition of a Brownian motion.
Now, it is known that for every random variable $$ Y: \Omega \to \mathbb{R}, $$ we have an induced probability measure $\mathbb{P}_Y$ in $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ given by
$$ \mathbb{P}_Y (B) := \mathbb{P}(Y^{-1}(B)), \quad \forall B \in \mathcal{B}(\mathbb{R}), $$ called the measure induced by $Y$.
My question therefore is if every probability measure is induced, that is:
Given a probability measure $\mu$ in $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, is it possible to find a random variable $$ Z:(\Omega, \mathcal{F}, \mathbb{P}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})) $$ such that $\mu = \mathbb{P}_Z$?
Thank you very much for you time!
Here are a couple of things worth knowing.
(1) If a random variable $Y$ has continuous cdf $F$, then $F(Y)$ is a random variable that is uniformly distributed on $[0,1]$. For example (related to Kabo Murphy's comment) if $\Phi$ is the standard normal cdf then $\Phi(X_1)$ is uniform on $[0,1]$. The proof amounts to calculating the cdf of $F(X)$.
(2) If $\mu$ is a probability measure on $(\Bbb R,\mathcal B(\Bbb R))$, define, for $0\le y\le 1$, $G(y):=\inf\{ x\in\Bbb R:\mu(-\infty,x]>y\}$, where $\inf\emptyset:=+\infty$. Now prove that if $U$ is uniformly distributed on $[0,1]$ then the distribution of $G(U)$ is $\mu$.
Combine (1) and (2) to answer your question (in the affirmative).