I'm actually working on brownian motion and continuous time stochastic process.(For the rest of the discussion we fix $(\Omega,\mathcal{F},\mathbb{P} )$ a probability space and all the objects are defined on this probability space). For Consequences and as you probably know a lot of subsets of $\Omega$ such that $\{\omega:X_t=Y_t\}$ are not necessarily measurable thus it makes no sense to look at $\mathbb{P}(\{\omega:X_t=Y_t\})$.
In order to avoid this problem I would like to define the outer measure generated by $\mathbb{P}$, let's call it $\tilde{\mathbb{P}}$ and $\tilde{\mathbb{P}}(E):=\inf \{\sum_{j=1}^\infty \mathbb{P}(A_j):E\subset \cup_{j=1}^\infty A_j, A_j \in \mathcal{F}\}$ for all $E\in 2^{\Omega}$.
My question is as follow. Do we have that if $\tilde{\mathbb{P}}(E)=0$, then $E$ is negligible. In particular it will allow me to prove that some statement holds a.s (here a.s is understood in the sense that the complement is negligible) without checking that the latter is measurable. Typically in computing $\tilde{\mathbb{P}}(\{\omega:X_t=Y_t\})$.
EDIT: To be clear, the probability space is not (a priori) complete.
Many thanks by advance for you help :)
Best,
Arthur
Here is a friend suggestion that looks nice:
given $E$ of outer measure $0$, lets consider for all $n\in \mathbb{N}$, $(A_j^n)_{j,n \in \mathbb{N}}$ such that $E\subset \cup_{j=1}^\infty A_j^n$ and $\sum_{j=1}^{\infty} \mathbb{P}(A_j^n)\leq \frac{1}{n}$ then consider $N:= \cap_{j,n \in \mathbb{N}} A_j^n$ which is measurable, contains $E$ and of measure $0$.