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.