$(\Omega,\mathcal{F},\mathbb{P})$ is our sample space. Consider two stochastic processes $X,Y$ that are almost everywhere equal (i.e., $\forall t, \mathbb{P}(X_t)=\mathbb{P}(Y_t)$ a.s.). Suppose that $X_t$ is adapted to a filtration $\mathcal{F}_t$, how to construct a counter-example in which $Y_t$ is not adapted to $\mathcal{F}_t$?
This problem is from this Annals-24 paper, which in pg.5 last paragraph states that "(the intensity process of a counting process) $\boldsymbol{\lambda}$ is not a prior $\mathcal{F}_t$ adapted but has a $\mathcal{F}_t$ adapted version".
I have discussed this with the author, but he cannot give a complete counter-example, either. I am aware of this post, but I (and the author of the paper) thought things might be different for stochastic process?
NOTE: for anyone who is new to measure theory (like myself), I explain the answer by @geetha290krm in detail.
Let $(\Omega,\mathcal{F},\mathbb{P})$, with $\mathcal{F}$ the Lebesgure $\sigma$-algebra and $\mathbb{P}$ the Lebesgure measure on [0,1]. Let $\mathcal{B}$ the Borel $\sigma$-algebra.
Recall that
$A$ is a Lebesgue measurable set if there is a Borel measurable set $B$ and a Lebesgue measure zero set $N$ s.t. $A=B\triangle N:=(B\backslash N)\cup(N\backslash B)$
Now, consider a Lebesgue measurable set $A$ (that is not Borel measurable). Let $B$, $N$ the corresponding two sets in the above definition. Let $X:=1_B, Y:=1_A$.
We can see that $\mathbb{P}(X\neq Y)=\mathbb{P}(N)=0$. Meanwhile, we have $X$ a Borel measurable function because $B$ is a Borel set, and $Y$ not Borel measurable because $A$ is not a Borel set. This means $X$ being measurable to $\mathcal{B}$ and $X=Y$ a.e. does not imply $Y$ being measurable to $\mathcal{B}$.
