# Almost everywhere measurable function definition

Given $(\Omega, \mathcal{M})$ measurable space and $(X,\tau)$ topological space.

A function $f: (\Omega, \mathcal{M}) \to (X,\tau) \$ is almost everywhere measurable if $\exists \ \Omega_0 \subset \Omega$ with $\mu(\Omega_{0}^{C}) = 0$ s.t. $\forall A \subset X$ open $f^{-1}(A) \cap \Omega_0 \in \mathcal{M}$

I am having some troubles in understanding this definition the lecturer gave me. I think it is pretty different from the measurable function's definition, i.e.

$\forall A \subset X$ open $f^{-1}(A) \in \mathcal{M}$

What if I say something like this:

A function $f: (\Omega, \mathcal{M}) \to (X,\tau)$ is almost everywhere measurable if $\exists \ \Omega_0 \subset \Omega$ with $\mu(\Omega_{0}^{C}) = 0$ s.t. $\forall A \subset X$ open $f^{-1}(A) \subset \Omega_0 \in \mathcal{M}$

Is there anything wrong with it? If so, why?

Let $g: \Omega_0 \to X = f|_{\Omega_0}$. Then notice that for $A \subset X$, $f^{-1}(A) \cap \Omega_0 = g^{-1}(A)$. So your lecturer's definition says that we can find a full subset $\Omega_0 \subseteq \Omega$ such that the restriction of $f$ to $\Omega_0$ is a measurable function.
Unfortunately, your definition is really not the same thing. In fact, your definition says essentially nothing. If $f$ is almost everywhere measurable according to your defintion then this implies that $\Omega_0 = \Omega$ since $X$ is always open and so we would have $\Omega = f^{-1}(X) \subset \Omega_0$. So if any $f$ satisfies your definition, then your definition just says that $f$ is a function from $\Omega$ to $X$ and nothing more!
Edit: @TheoreticalEconomist pointed out in the comments that your definition may not parse the way you wanted it to. If you intended to require that $f^{-1}(A) \in \mathcal{M}$ then the fact that $\Omega_0 = \Omega$ means that your definition is nothing other than the definition of measurability!