In light of the comments here, I'm going to show why completeness can be a pain. In exercise 9 of section 2.1 of Folland, he develops a function $g: [0,1] \to [0,2]$ by $g(x) = f(x) + x$ where $f : [0,1] \to [0,1]$ is the Cantor function. In that exercise it is established that $g$ is a (monotonic increasing) bijection, and that its inverse $h = g^{-1}$ is continuous from $[0,2]$ to $[0,1]$.

Since $h$ is continuous, it is Borel measurable. On the other hand, $h$ is not $(\mathcal{L}, \mathcal{L})$-measurable!! In particular, let $C$ be the Cantor set; $m(g(C)) = 1$, but this means there is a subset $A \subseteq g(C)$ which is not Lebesgue measurable. On the other hand $B := g^{-1}(A) \subseteq C$ whereas $m(C) = 0$; thus this preimage $B$ is Lebesgue measurable (with measure zero). But therefore $h^{-1}(B) = A$ is not Lebesgue measurable, meaning $h$ is not $(\mathcal{L}, \mathcal{L})$-measurable.

On one hand, this function is contrived. On the other hand, it shows that completing measures can mess things up. The typical definition of "measurable function" is a Borel
measurable function, and I suppose reasons like the above led to this convention. I do not know the material Bridge references above, and so can't say what breaks when completeness is dropped. Although it seems mathematically convenient to throw in completeness, I don't know any examples in basic probability theory where it helps. For instance, Fubini-Tonelli can be formulated just fine without completeness. Your statement of the theorem only need mention completeness if your measures happen to be complete!

**EDIT** I corrected the nonsense in the second paragraph; also I meant to talk about $(\mathcal L, \mathcal L)$-measurable functions, which I accidentally refered to as Lebesgue measurable (which means $(\mathcal L, \mathcal B)$-measurable). My whole point is that if you take completion in $\sigma$-algebra of the range space, the extra sets you added could map back to basically anything. IE it is somewhat nonsensical to add in all sorts of null sets, but not all sorts of finite measure sets. Sometimes completion gives you something you want, but sometimes it does not, as I showed here--the function is better behaved wrt the non-completed measure.