One should be careful with the definitions here. Notation: Given measurable spaces $(X, \mathcal{B}_X), (Y, \mathcal{B}_Y)$, a measurable map $f : X \to Y$ is one such that $f^{-1}(A) \in \mathcal{B}_X$ for $A \in \mathcal{B}_Y$. To be explicit, I'll say $f$ is $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable.
Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\bar{\mathcal{B}}$ is its completion. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\bar{\mathcal{B}},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)
$(\bar{\mathcal{B}}, \bar{\mathcal{B}})$-measurable functions are not so useful; for instance, a continuous function need not be $(\bar{\mathcal{B}}, \bar{\mathcal{B}})$-measurable. (The $g$ from the aforementioned exercise is an example.)
Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \bar{\mathcal{B}})$-measurable functions; there simply may not be enough good ones. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.
Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \bar{\mathcal{B}})$ in our definition of "random variable" would not avoid this, either.
I think the moral is that a complete $\sigma$-algebra is nice to have on the domain of a measurable function. It is not necessarily so nice to have on the codomain.