Is $\max(f)$ well defined if $f$ is finite? Say I have a function that's finite almost everywhere within an interval $[a,b]$. Does that mean that it has an upper bound if I ignore those points on which it is infinite? i.e. is: 
$$\text{max}_{[a,b]}(f)$$
well defined?
 A: No: Consider $f(x) = \frac{1}{x}$ if $x > 0$, and $f(0) = 0$. This is finite everywhere on $[0, 1]$, but has no meaningful maximum.
Your question seems related to the essential supremum, which is defined to be the least $\alpha$ for which 
$$|f(x)| \le \alpha$$
for almost every $x \in [a, b]$. But there are certainly functions (such as the one I described above) which are finite everywhere but are neither bounded nor essentially bounded.
A: No, for one of two possible reasons:
1) The function may still be unbounded in $[a,b]$. Bounded and finite are not the same concept. Bounded means there is a number $M<\infty$ such that $|f(x)|<M$ for $x\in[a,b]$, whereas finite means that $|f(x)|<\infty$ for $x\in[a,b]$. This doesn't change if you say these properties hold almost everywhere.
2) Even if the function is finite a.e. in $[a,b]$, it may not achieve its supremum, in which case it has no maximum. There are conditions on $f$ that will guarantee it achieves its supremum (e.g. defined everywhere on a compact set and continuous), but under your minimal assumptions this cannot be guaranteed.
However, in the case that $f$ is unbounded on $[a,b]$ minus a null set and achieves $\infty$ on this null set, then if you allow yourself the extended real numbers you can take $\max_{[a,b]} f = \infty$.
