How to prove the function $g=0$, $\mu$-a.e is a measurable function Let $(\Omega, \Sigma, \mu)$ be a measure space. Let $(\mathbb{R}, \mathcal{B})$ be the measurable space. Let $g:\Omega\rightarrow \mathbb{R}$ be a function such that
$$g=0,\ \mu-\mbox{a.e}.$$
One thing clear here is that
$$\mu(\{\omega\in\Omega: g(\omega)\neq 0\})=0.$$
Now, I have to prove that the function $g$ is measurable. In order to do that, I have to show that for an arbitrary $x\in\mathbb{R}$, we have that
$$\{\omega : g(\omega)\leq x\}\in \Sigma.$$
Now, let $x\in(-\infty,0)$, then the set $\{\omega : g(\omega)\leq x\}$ is not necessarily
empty because the thing $\mu-$a.e. Then how can it be shown that
$$\{\omega : g(\omega)\leq x\}\in \Sigma.$$
 A: Let $\Omega = \{1,2,3\}$ and $\Sigma =\{\emptyset,\{1\},\{2,3\},\Omega\}$ and $\mu=\delta_1$. Let $u(k)=k-1$. Then $u=0$ $\delta_1$-a.e. because $\delta_1(\{k:u(k) \neq 0\})=\delta_1(\{2,3\})=0$. However, $u^{-1}(\{1\})=\{k:u(k)=1\}=\{2\}\notin \Sigma$. Therefore $u$ is not measurable.
A: Measurability of a function $g$ relies on both $\sigma$-algebras of domain and values, here $\Sigma$ and $\mathcal{B}(\mathbb{R})$. For example you can set $\mu=0$ and every function $g$ will satisfy $g=0$ $\mu-a.e.$. On other side you could set $\Sigma=\mathcal{P}(\Omega)$ and every function $g$ will be measurable.
Therefore you need more conditions on the sigma algebras to conclude, when a function $g$ is measurable or not.
A: Both previous answers are absolutely right. Your objects are not perfectly well-defined in the sense that you cannot define a function almost surely without knowing it's measurable beforehand for the reasons said above (in the previous questions).
Something you could wonder however is:
Is the constant map $f:\Omega\to \mathbb R,\omega\to 0$ measurable? And the answer is yes no matter the structure of the $\sigma$-algebra $\Sigma$ since we have
$$
f^{-1}(B)\in \{\emptyset ,\Sigma\},\quad B\in \mathcal B(\mathbb R).
$$
