Dominated Convergence Theorem - need complete measure? Here is the theorem I was reading: (p54, Folland)

$(X,\mathcal{M},\mu)$ is a measure space. Let $\{f_n\}$ be sequence in $L^1(\mu)$ such that $f_n \rightarrow f$ a.e and exists $g \in L^1(\mu)$ such that $|f_n| \le g$ for all $n$. Then $f \in L^1 (\mu)$ and $\int f = \lim \int f_n $.

How do we know $f$ is measurable? We know $g:=\limsup f_n$ is measurable, and $g=f$ a.e. But for $f$ to be measurable do we not need complete measure?

This is a standard abuse as others have pointed out.
You should think of defining $f$ to be equal to the pointwise limit $\lim_{n \to \infty} f_n(x)$ wherever that limit exists and equal to anything you like, such as 0, when that limit does not exist. This will be a measurable function.
Or if you can stand it, $f$ is only defined almost everywhere. But then $f^{-1}(B) = \{x : f(x) \in B\}$ for a Borel set $B$ can only possibly be a subset of the set of points where $f$ is defined.
