To what categorical concept is this proposition equivalent to... In measure theory we have the following result (I know it's true for positive valued functions, so I'm taking a leap of faith assuming it's true for any measure space.)
Proposition Let $(X,\overline{M},\overline{\mu})$ be the completion of $(X,M,\mu)$ and $f:(X,\overline{M},\overline{\mu})\to (Y,N,\nu)$ a measurable function. Then, there exists a measurable function $g:(X,M,\mu)\to (Y,N,\nu)$ such that $f=g$ $\overline{\mu}$-a.e.
So, I'd like to know if it's possible to formulate the proposition in terms of a commutative diagram in the category of measure spaces. Something like: 
Given $f$ measurable there exists $g$ measurable such that the following diagram commutes:

And that the commutativity of the diagram implies the equality almost everywhere, i.e., $$f = g\circ (1_X?) \iff f=g\text{ $\overline{\mu}$-a.e.}$$
Thinking about this led me to the following more general questions:
$(1)$ Is it possible, in general, to define equality almost everywhere in terms of commutative diagrams?
$(2)$ The identity gives us an inclusion (an injective measurable function) of the completion into the original space, does this mean that the completion is "smaller" than the original space?
If this is the case, this proposition is basically an "extension" property, so
$(3)$ what else can be said about this? Is this "extension" property a particular example of a categorical construction? 
The diagram and property reminds me about the universal property of a free object, could it be that $(X,M,\mu)$ is the free object generated by $(X,\overline{M},\overline{\mu})$?
 A: *

*Note that if $f,g: (X,M,\mu) \to (Y,N,\nu)$, then to say that $f = g$ $\mu$-a.e. is to say that the complement of the equalizer of $f$ and $g$ (in the category $\mathsf{Meas}$ of measure spaces and measurable maps) has $\mu$-measure-0. To say that $f=g$ $\bar\mu$-a.e. is to say that the complement of the equalizer of $f$ and $g$ is contained in a set of $\mu$-measure 0.

*The map $"1_X" (X,\bar M, \bar \mu)  \to (X,M, \mu)$ is the identity on points, so I wouldn't say that it means the completion is "smaller" than the original space. It's analogous to passing from a finer topology to a coarser topology on a space. One interesting fact is that by taking $f$ to be the identity on $(X, \bar M, \bar \mu)$, one obtains a retraction of $"1_X"$ up to a.e. equivalence.

*I think you're right that this smells like an adjunction. It's a little strange, though: your diagram would look like it was saying that the completion process was left adjoint to the inclusion of the category of complete measure spaces into the category of all measure spaces, except that you would need to swap the places of $X$ and its completion for this to work.
Also, there's this business of needing to say "up to a.e. equivalence" everywhere. One would like to solve this quotienting by the a.e.-equivalence relation, but unfortunately this doesn't work because a.e.-equivalence is not a congruence relation on the category of measure spaces and measureable maps (and if you quotient by the smallest congruence containing it, you get something equivalent to the terminal category or perhaps the arrow category if you include the empty space).
We can address this by passing to a subcategory. Let $\mathsf{Meas}'$ be the category whose objects are measure spaces and whose morphisms are measurable maps $f: X \to Y$ such that if $V \subseteq Y$ has measure 0, then $f^{-1}(V)$ has measure 0. Then a.e. equivalence (in the sense of $f=g$ a.e. if $\{x \mid f(x) \neq g(x)\}$ is contained in a set of measure 0) is a congruence relation on $\mathsf{Meas}'$, and we can quotient by it to get a category $\mathsf{Meas}'_{a.e.}$.
Your proposition amounts to the observation that in $\mathsf{Meas'}_{a.e.}$, every space is isomorphic to its completion. This sets up an equivalence of categories between $\mathsf{Meas}'_{a.e.}$ and the full subcategory complete measure spaces, which can be improved to an adjoint equivalence in a few ways.
There may be other, more elegant categorical analyses of this situation though.
