Let $\cal B$ be the set of all Borel-measurable subsets of $\Bbb R$ and $\cal L$ be the set of all Lebesgue-measurable subsets of $\Bbb R$. In measure theory texts, a function $f:\Bbb R\to \Bbb R$ is said to be measurable if for every $B\in \cal B$, we have $f^{-1}(B)\in \cal L$.
Why is $\cal B$ used for range of the function. Why $\cal L$ is not used on both sides:
a function $f:\Bbb R\to \Bbb R$ is said to be measurable if for every $L\in \cal L$, we have $f^{-1}(L)\in \cal L$
?
The basic reason is that the primary purpose of measurable functions is being able to integrate them, and we only need Borel sets in the codomain to define integrals. Indeed, you can define the integral of a measurable function by approximating it by simple functions. For these simple functions to be measurable, you need sets of the form $f^{-1}(I)$ to be measurable, where $I\subseteq\mathbb{R}$ is an interval. The $\sigma$-algebra generated by intervals is the Borel $\sigma$-algebra, so asking for $f^{-1}(I)$ to be measurable for all intervals $I$ is equivalent to asking for $f^{-1}(B)$ to be measurable for all Borel $B$.
So, for most purposes, we don't lose anything if we only require the inverse images of Borel sets to be measurable, rather than the inverse sets of Lebesgue measurable sets. On the other hand, we do lose quite a lot if we use Lebesgue measurable sets on the codomain. In particular, not every continuous function (or even every homeomorphism) $\mathbb{R}\to\mathbb{R}$ would be measurable, if we used $\mathcal{L}$ on both the domain and codomain. (Proof sketch: take a fat Cantor set $C$ and the usual null Cantor set $D$, and a homeomorphism $f:\mathbb{R}\to\mathbb{R}$ that maps $C$ to $D$. Every subset of $D$ is in $\mathcal{L}$ since it is null, but not every subset of $C$ is in $\mathcal{L}$, so $f$ is not $\mathcal{L}$-to-$\mathcal{L}$ measurable.)
Underlying all of this is that the domain and codomain play fundamentally different roles when we're talking about measurable functions $\mathbb{R}\to\mathbb{R}$. The domain is a measure space: we care about measuring the size of subsets of this, and using this to more generally define integrals of functions on it. We don't actually care that the domain is $\mathbb{R}$ specifically; we can make similar definitions for functions $X\to\mathbb{R}$ where $X$ is any measure space. So, that's the difference between using $\mathcal{B}$ and $\mathcal{L}$ on the domain: we're just changing what our domain measure space is.
On the other hand, the codomain $\mathbb{R}$ is all about actual real numbers and their algebraic (and topological) properties. An integral of a real-valued function is a kind of "continuous weighted sum" of values of the function, so we care very much about being able to take linear combinations of elements of our codomain, and take limits of them. The natural generalization of the codomain is to consider an arbitrary topological vector space, rather than an arbitrary measure space. We don't care about measuring sizes of subsets of our codomain, and so we don't care about throwing in arbitrary Lebesgue null sets.