Why not: Given a measurable space $(\Omega, \mathcal{A})$, a function $X:\Omega \to \mathbb{R}$ with Borel $\sigma$-algebra $\mathfrak{B}(\mathbb{R})$, is measurable if $\forall A \in \mathcal{A}, X(A) = B \in \mathfrak{B}(\mathbb{R})$ ?
I'm confident that my version is wrong somehow---very smart people don't write definitions casually---but for the life of me I can't see why my (simpler?) version would not work. Assuming that I'm right about being wrong, a simple example where it doesn't work would be better than a technical explanation as I'm very much a novice.
With your definition the characteristic function of any set would be measurable, since the image $\{1\}$ is Borel.
Now, working with the usual Vitali non-measurable set $E$ , you have a set $E\subset[0,1] $ such that $$ [0,1]=\bigcup_r (E+r), $$ where $r$ runs over the rationals in $(0,1)$.
Now, the whole point of measure theory is to define integration. An integral should be linear and, when trying to generalize Riemannian integration, translation invariant. Also, the spirit of an integral is that for a positive function on the real line it should be "the area under the curve" or, more generally, of the form "value of the function times size of the region". So you would want $$\tag1 \int_{\mathbb R}1_{[0,1]}=1=\sum_r\int_{\mathbb R}1_{E+r}. $$ The problem is that if the integral is going to be translation invariant, then all integrals on the right should be equal; if they are nonzero then the sum is infinite; and if they are zero then the sum is zero. So there is no way to make the equality $(1)$ to work.