Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $W:[0,\infty)\times \Omega\rightarrow\mathbb{R}$ be a standard Brownian motion. Is it true that for all $x\in\mathbb{R}$ there exists $\omega\in\Omega$ such that $W(1,\omega)=x$?
I do not know if the question has a positive or negative answer. I know that $P(W(1)\in [a,b])>0$ for all real numbers $a<b$, but I do not think that suffices. I have no idea on how to prove the existence of such an $\omega$, since it seems to me that the proof may use an explicit form of $\Omega$.
EDIT: As I read in a comment below, this statement is not true. My doubt came from reading the proof of Lemma 5.22 in An Introduction to Computational Stochastic PDEs. The step I do not completely understand is the following: from $E[W(t)|W(1)]=t \,W(1)$, where $W$ is a Brownian motion and $0\leq t\leq 1$, it is stated that $E[W(t)|W(1)=0]=0$.
One of the possible definitions that I studied for $E[W(t)|W(1)]$, and in general for $E[X|Y]$, where $X$ and $Y$ random variables, is $E[X|Y](\omega):=E[X|Y=Y(\omega)]$, where $E[X|Y=y]=\int_{\mathbb{R}} x\,P_{X|Y=y}(dx)$, being $P_{X|Y=y}$ the $P_Y$-unique probability satisfying $P(X\in A,Y\in B)=\int_B P_{X|Y=y}(A)\,P_Y(dy)$.
In this example, $E[W(t)|W(1)=W(1)(\omega)]=E[W(t)|W(1)](\omega)=t\,W(1)(\omega)$. I understand that, if $W(1)(\omega)=0$ for some $\omega\in\Omega$, then $E[W(t)|W(1)=0]=t\cdot 0=0$. But if for our particular probability space $(\Omega,\mathcal{F},P)$ and our particular Brownian motion $W$ there is no $\omega$ satisfying $W(1)(\omega)=0$, then I do not see that $E[W(t)|W(1)]=t\,W(1)$ implies that $E[W(t)|W(1)=0]=0$.
Could you explain this to me using probability theory, and not just "intuition"?