almost sure convergence for non-measurable functions

Question

Let $(\Omega,\mathscr{F},P)$ be a probability space. Assume for each $n$, $Y_n:\Omega\rightarrow\mathbb{R}$ is a function but $Y_n$ is not necessarily $\mathscr{F}$-measurable. In this case, is it still meaningful to talk about almost sure convergence of $Y_n$? Conceptually, we can define almost sure convergence as $$\exists\hat{\Omega}\in\mathscr{F}\quad\mathrm{such\,\,that}\quad P(\hat{\Omega})=1\quad\mathrm{and}\quad \{Y_n(\omega)\}\,\,\mathrm{converges}\,\,\forall \omega\in\hat{\Omega}.$$

In every probability textbook I have, they all define almost sure convergence for "random variables". But I think what I mentioned might arise naturally in some situations. For example, if for each $n$, $\{X^n_\lambda\}_{\lambda\in \Lambda}$ is a class of random variables where $\Lambda$ is uncountable, then $$Y_n\equiv\sup_{\lambda}X_\lambda^n$$ is not necessarily measurable, but still we sometimes want to talk about convergence property of $\{Y_n\}$.

Answer 1

Not only a.s. convergence but pointwise convergence, as well, can be defined in the case of sequences of non measurable functions. Let, for instance, $$([0,1],\mathscr A=\left\{\emptyset,[0,1/2],(1/2,1],[0,1]\right\},\mathbb P((0,1/2]))=\mathbb P((1/2,1])=1/2)$$ be a probability space, and let $$X_n(\omega)=\frac{\omega}{n}, \text{ if }\ \omega\in[0,1].$$ Obviously $X_n$ converges pointwise to $0$ on $[0,1]$. So far so good. However, there is no answer to important$^*$ questions. Consider only the following example: $$\mathbb P\left(X_3<\frac{1}{5}\right)=\ "\mathbb P"\left(\left\{\omega:0\le \omega<\frac{3}{5}\right\}\right)=??$$ There is no answer because $\mathscr A$ and $\mathbb P$ could be extended many different ways.

$^*$ Philosophcal-BTW: What is important at all?