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 $\mathcal{L}$ is its completion with respect to Lebesgue measure $m$. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\mathcal{L},\mathcal{B})$ 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.)
$(\mathcal{L}, \mathcal{L})$-measurable functions are not so useful; for instance, a continuous function need not be $(\mathcal{L}, \mathcal{L})$-measurable. (The $g$ from the aforementioned exercise is an example.) $(\mathcal{B}, \mathcal{L})$ is even worse.
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 $\mathcal{L}$ does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \mathcal{L})$-measurable functions; there simply may not be enough good ones, and they may not be preserved by composition with continuous functions. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$ with respect to $P$, 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 $(\mathcal{F}, \mathcal{L})$ in our definition of "random variable" would not avoid this, either.
The moral is this: To get as many $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra there. (You already know some of the nice properties of this, e.g. an a.e. limit of measurable functions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. When $Y$ is a topological space, we usually want to be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the open sets (and hence the Borel $\sigma$-algebra), but we should stop there.
Best Answer
Yes, since a convex function $f:\mathbb{R}^d\rightarrow\mathbb{R}$ is locally Lipschitz (see e.g. Theorem 6.3.1, pp. 236-239 of the book by L. C. Evans and R. F. Gariepy, Measure Theory and Fine Properties of Functions, CRC Press, 1992).
Recall that, as you wrote it yourself in the particular case of convex functions and more generally by Rademacher's theorem, if $g:\mathbb{R}^d\rightarrow\mathbb{R}$ is locally Lipschitz then $g$ is differentiable $\lambda^d$-almost everywhere. Moreover, in that case the distributional partial derivatives of $g$ are in $L^\infty_{loc}(\mathbb{R}^d,\lambda^d)$ and can be identified $\lambda^d$-almost everywhere (say, on a Borel subset $A\subset\mathbb{R}^d$ with $\lambda^d(\mathbb{R}^d\smallsetminus A)=0$ as you defined it) with the classical partial derivatives of $g$ (for the details of all statements above for locally Lipschitz functions, see e.g. Theorem 5 in subsection 4.2.3, pp. 131-132 and Theorems 6.2.1, 6.2.2, pp. 235 of Evans-Gariepy), hence these partial derivatives are in $L^\infty_{loc}(A,\lambda^d)$ and therefore are (Borel) measurable.