Good question; I didn't know the answer until just now, never having thought about it.
A: Because if we did there would be fewer measurable functions, and in particular a continuous function $f:\Bbb R\to\Bbb R$ would not necessarily be measurable:
Say $K$ is a "fat Cantor set", that is a subset of $\Bbb R$ homeomorphic to the middle-thirds Cantor set $C$ but such that $m(K)>0$. There is a continuous bijection $f:\Bbb R\to\Bbb R$ such that $f(K)=C$. Now $K$ contains a non-measurable set $E$. Say $F=f(E)$. Then $F$ is Lebesgue measurable, being a subset of a null set, but its inverse image $E$ is not Lebesgue measurable. So $f$ is continuous but not Lebesgue-to-Lebesgue measurable.
Not what we want; if continuous functions are not measurable none of the things we want to use this stuff for work anymore.
(Also it's fun to ask students to resolve the "paradox" that according to the definition of a measurable function between measurable spaces, no topology, the composition of two measurable functions is obviously measurable, yet in this context the composition of two measurable functions need not be measurable.)