Let $f:\mathbb R\to \mathbb R$ a function. Why $f$ is measurable if $f^{-1}(B)\in \mathcal M$ for all $\mathcal B$ and is not $f^{-1}(M)\in \mathcal M$ for all $M\in \mathcal M$ ? (where $\mathcal M$ is the $\sigma -$algebra of Lebesgue measurable set and $\mathcal B$ the $\sigma -$algebra of Borel set ?
What would be the problem if we would define a measurable function by $f^{-1}(M)\in \mathcal M$ for all $M\in\mathcal M$ ?
Consider a function $f:(\Bbb R,\mathcal A_1)\to (\Bbb R,\mathcal A_2)$, where $\mathcal A_i$'s are the $\sigma$-algebras on $\Bbb R$. We want nice functions to be measurable and I hope you'd agree that continuous functions should be considered nice.
The $\sigma$-algebra we should impose on $\mathcal A_2$ (the target space) that is compatible with the usual topology on $\Bbb R$ is the Borel $\sigma$-algebra $\mathcal B$ since it is the $\sigma$-algebra generated by $\tau_{\Bbb R}$, the topology of $\Bbb R$.
On the other hand, while it is possible to let $\mathcal A_1=\mathcal B$ (this is enough if we merely require that all continuous functions be measurable), it is more convenient if we can say that "all subsets of null sets are null" because the adjective "almost everywhere" is linked to the $\sigma$-algebra $\mathcal A_1$. Hence, we usually take the Lebesgue $\sigma$-algebra $\mathcal A_1=\mathcal M$, the completion of $\mathcal B$, as the $\sigma$-algebra on our domain.
PS. Here's my answer to a related question regarding why choosing $\mathcal A_1 = \mathcal A_2 = \mathcal B$ is sometimes better than $\mathcal M-\mathcal B$.
The definition of $f:\mathbb R\to \mathbb R$ is a Lebesgue measurable function is that $\{x\mid f(x)\leq a\}$ is Lebesgue measurable for all $a\in\mathbb R$. And this is equivalent to $f^{-1}(B)\in \mathcal M$ for all Borel set $B$. Now, in more general way, if $(X,\mathcal A)$ and $(Y,\mathcal C)$ are measure space, then $$f:(X,\mathcal A)\to (Y,\mathcal C)$$ is measurable if $f^{-1}(C)\in \mathcal A$ for all $C\in \mathcal C$. So if you consider a function $f:(\mathbb R,\mathcal M)\to (\mathbb R,\mathcal M)$, then you'll have that $f$ is measurable if $f^{-1}(M)\in \mathcal M$ for all $M\in \mathcal M$.
Now, the idea behind function Lebesgue measurable is to extend a bit the class of continuous function. A natural way to extend this class is to consider function such that $f^{-1}(O)$ is Lebesgue measurable for all open $O$ (it's equivalent to the definition I gave above, i.e. to $\{x\mid f(x)\leq a\}\in \mathcal M$ for all $a\in\mathbb R$) instead of $f^{-1}(O)$ open for all open set $O$. If you consider measurability of function as $f^{-1}(M)\in \mathcal M$ for all $M\in\mathcal M$, then very pathological functions will arise... we therefore want to avoid it. But after all, why not ?
