Does changing the values of a $\mathbb P$-a.s. defined map at countably many points preserve the map? Let $(\Omega,\mathcal F,\mathbb P)$ be a non-atomic probability space and $X:\Omega\rightarrow\mathbb R^d$ a $\mathcal F$-measurable map. Let $(\omega_k)_{k\in\mathbb N}\subseteq\Omega$ and $(y_k)_{k\in\mathbb N}\subseteq\mathbb R^d$ be two sequences. I want to define a $\mathcal F$-measurable map $\tilde X$ such that $$\mathbb P(X=\tilde X)=1$$ and 
$$\tilde X(\omega_k)=y_k$$ for some $(\omega_k)_{k\in\mathbb N}\subseteq\Omega.$
My idea is that non-atomicity should allow to change the value of $X$ at countably many points, since countably many points have $\mathbb P$-measure $0$. But does it still preserve measurability and is the map then still the same $\mathbb P$-a.s.?

First, every countable set in an atomless probability space has outer measure zero, so if the function is measurable, it agrees with the old one almost surely.
But the new function may not be measurable. Let $\Omega=[0,1]^2$ $\mathcal{F}=\{B\times [0,1]: B$ a Borel subset of $[0,1]\}$ and $P$ be defined so that $P(B)$ is the Lebesgue measure of the projection of $B$ onto its first coordinates. Let $f$ be constant and change it on a single point. Since points are not measurable (the smallest nonempty measurable sets are lines ${x}\times[0,1]$), the resulting function is not measurable.
This problem can not occur when points are measurable. This is for example the case when $P$ is complete or when the underlying space is a Hausdorff space with the Borel $\sigma$-algebra.
