Is the empty set Lebesgue measurable?

Measures are defined on $\sigma$-algebras (containing the measurable sets) and any $\sigma$-algebra contains the empty set by definition.

Intuition

Given a set $X$, to avoid paradoxes such as the Banach Tarski paradox , there is a need to define which subsets are measurable (and as a consequence which ones aren't), and then define a function $\mu$, called a measure, from this set $\Sigma$ of subsets of $X$ to the non-negative reals. How to choose?

You want the measure to mirror our intuition of volume, or length, generalising it to higher dimensions and/or to non-metric spaces. In particular, you want the whole set $X$ to have a measure, so that $X \in \Sigma$. Another property you want is that if a subset $A\in \Sigma$ has a measure, you want its complement $X/A$ to have measure $\mu(X)-\mu(A)$. These two properties are necessary (but not sufficient) for $(X,\Sigma,\mu)$ to be a measure space, i.e. as space that generalises our intuition of volume without logical inconsistencies.

Answer

For any measure space $(X,\Sigma,\mu)$, the empty set $X / X \equiv \emptyset$ needs to be a measurable set, and its measure is invariably $\mu(X)-\mu(X) = 0$.

Any $\sigma$-algebra contains $\emptyset$ and for any measure $\mu$ on the $\sigma$-algebra, $\mu(\emptyset) = 0$.