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, 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

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, 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 open left-halves of the omitted intervals, and let $R$ be the union of the open right-halves of those intervals. So $L$ and $R$ form a disjoint open partition of $U$, minus the countably many center points of the omitted intervals, 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 of the 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

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, obtained by omitting much less than the middle third of each of the remaining intervals. For example, if you omit the middle sixths at each step, then you get a Cantor set of measure $1/2$, or if you omit proportion $\epsilon/3$, then the Cantor set will have measure $1-\epsilon$.

Let $U$ be the union of those omitted intervals, the complement of the fat Cantor set. This is an open dense set of measure $\epsilon$  (and so $U$ of course, is not regular open).

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

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, 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