3 no idea how i hallucinated the mistakes, but they are now gone

Since $h$ is continuous, it is Borel measurable. On the other hand, $h$ is not Lebesgue measurable!! $(\mathcal{L}, \mathcal{L})$-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 $g^{-1}([0,2]) B := g^{-1}(A) \subseteq C$ and whereas $m(C) = 0$; therefore the pre-image of any subset of $[0,2]$ (and in particular thus this preimage $A$) B$is a subset of a Lebesgue measure-zero set, and thus it is a measurable set(with measure zero). This provides that But therefore$h$h^{-1}(B) = A$ is not Lebesgue measurable., meaning $h$ is not $(\mathcal{L}, \mathcal{L})$-measurable.

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.

Post Undeleted by Matus Telgarsky
2 rewrote, original needed to be a little more careful..

In light of the comments here, I'm going to play devil's advocate and point out that completion show why completeness can actually be a pain. Take two Borel measurable functions In exercise 9 of section 2.1 of Folland, he develops a function $f,g$; then g: [0,1] \to [0,2]$by$g(x) = f(x) + x$where$f : [0,1] \circ to [0,1]$is the Cantor function. In that exercise it is established that$g$is also a (monotonic increasing) bijection, and that its inverse$h = g^{-1}$is continuous from$[0,2]$to$[0,1]$. Since$h$is continuous, it is Borel measurable. On the other hand, if$f,g$are h$ is not Lebesgue measurablemeasurable!! In particular, it let $C$ be the Cantor set; $m(g(C)) = 1$, but this means there is not true that a subset $f A \circ g$ subseteq g(C)$which is not Lebesgue measurable! A concrete example is given as exercise 9 . On the other hand$g^{-1}([0,2]) = C$and$m(C) = 0$; therefore the pre-image of any subset of$[0,2]$(and in section 2.1 particular$A$) is a subset of Follanda Lebesgue measure-zero set, where and thus it is even further provided that$g$is continuous!! This is a big deal because, especially in probability theory, measurable functions are one of the basic objects (random variables)set. That the class This provides that$h\$ is not closed under composition would make for a disasterLebesgue measurable.As such, when

On one says a hand, this function is measurablecontrived. On the other hand, it actually means it shows that completing measures can mess things up. The typical definition of "measurable function" is a Borelmeasurable function, and I suppose reasons like the above led to this convention. Fans of topology will also note that I do not know the Borel class in general material Bridge references above, and so can't say what breaks when completeness is quite nicedropped.

I'll say further as an anecdote that all the places Although it seems mathematically convenient to throw in completeness, I don't know of any examples in basic probability theory where completion it helpsare perhaps of mathematical interest. For instance, but not of real help to measure theoryFubini-Tonelli can be formulated just fine without completeness. Unfortunately I do not have Your statement of the background to comment on Bridge's answer in stochastic processes above, perhaps things go very wrong theorem only need mention completeness if completion is not taken.your measures happen to be complete!

Post Deleted by Matus Telgarsky
1