This is a great question! But unfortunately, the answer is no,  the Lebesgue measure on the unit
interval is not finitely $\vee$-additive.

**Theorem.** There are two disjoint regular open sets $L$ and $R$
in the unit interval, with Lebesgue measure as small as desired,
but whose union is dense, and so $L\vee R$ has full measure.

**Proof.** Consider the construction of a [fat Cantor set](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), obtained
by successively omitting much less than the middle third of each of the
remaining intervals. By omitting less, you can arrange that the resulting Cantor set has measure as close to $1$ as desired. 

Let $U$ be the union of those omitted intervals, the complement of
the fat Cantor set. This is an open dense set of some measure $\epsilon$, as small as desired. (The set $U$, being open dense, is not itself regular.) 

Let $L$ be the union of the left-halves of the omitted intervals,
and let $R$ be the union of the right-halves of those intervals. So
$L$ and $R$ form a disjoint open partition of $U$, and the measure
of $L$ and $R$ are each $\epsilon/2$.

I claim that each of $L$ and $R$ are regular open sets. To see
this, notice first that between any two intervals used to construct
$L$, there is an interval of $R$, and vice versa. Suppose that $u$
is an open interval contained in the closure of $L$. Since $R$ is
open and disjoint from $L$ and hence also from the closure of $L$,
it must be that $u$ contains no points from $R$. It follows that
$u$ can contain points from at most one interval of $L$, and from
this it follows that $u\subset L$. So the interior of the closure
of $L$ is $L$ itself and therefore $L$ is regular open. A similar
argument shows that $R$ is regular open.

Meanwhile, since the union $U=L\cup R$ is open dense in the unit
interval, it follows that $L\vee R$ is the whole interval, and so
the measure of $L\vee R$ is $1$.

So we have $\lambda(L)+\lambda(R)=\epsilon<1=\lambda(L\vee R)$,
which violates finite $\vee$-additivity. **QED**