# Almost everywhere limit of Borel measurable scalar functions is Borel measurable in a complete measure space

I am trying to solve the following exercise from Royden:

Let $(\Omega, \Sigma, \mu)$ be a complete measure space and $f_{n}:\Omega\to\mathbb{R}$ be measurable for each $n\geq 1$. If $f_{n}\to f$ almost everywhere on $\Omega$, then $f$ is measurable.

What I was thinking might work is if I could somehow construct an increasing sequence of measurable functions $g_{n}:\Omega\to\mathbb{R}$ which also converge to $f$ almost everywhere. That is, $g_{n}(\lambda)\geq g_{n-1}(\lambda)$ for all $\lambda\in\Omega,n\geq 1$.

Then I could use the fact that $\{\lambda\in\Omega : f(\lambda) \geq \alpha\}$ is "almost equal" to $\bigcup_{n=1}^{\infty}\{\lambda\in\Omega : f_{n}(\lambda) \geq \alpha\}$, and pull some magic with completeness to get that $\{\lambda\in\Omega : f(\lambda) \geq \alpha\}\in\Sigma$.

Is there a way to construct such $g_{n}$?

The proof is the same as the case of $(\mathbb{R},\mathcal{B},\mu)$, where $\mu$ is the Lebesgue measure.