Prove that $\,\lambda^*(Α)+\lambda^*(Β)= \lambda^*(A\cup B)\,$ if $\,A\subseteq G\,$ and $\,G\cap B=\emptyset\,.$

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What could be a rigorous proof that $λ^{*}(Α) + λ^{*}(Β) = λ^{*}(A\cup B)$ ?

We know that $A\subseteq G$ and $G\cap B = \emptyset$, and $\,λ^*$ the Lebesgue outer measure. Suppose all sets involved are measurable and subsets of $\,\mathbb R$.

One proof might be to use the definition of the measure, and the fact that $A\cup B=(a,b)\cup(c,d)$ where $(a,b)$ and $(c,d)$ the smallest possible sets in $\mathbb R$ containing $A$ and $B$ respectively.

But I don’t know if the professor would accept that or if there is a more rigorous way to prove this equality (e.g., one based on some property of the Lebesgue measure)