Q:Let and be given sets of finite outer measure. Show that m*( ∪ ) = m*(A) + m*(B) if and only if there are measurable sets A1 and B1 such that ⊂ A1 , ⊂ B1 and (A1 ∩ B1) = 0.Here Measurable sets means Lebesgue measure sets and m* means lebesgue outer measure.
I have proved the converse part.i just don't know how to approach the forward part, that is Given that m*( ∪ ) = m*(A) + m*(B) . Then To Prove:There are measurable sets A1 and B1 such that ⊂ A1 , ⊂ B1 and (A1 ∩ B1) = 0.
Suppose more generally that $\mu$ is a measure on $\mathcal{F}=\sigma(\mathcal{A})$, where $A\in\mathcal{A}$ is a subset of $\Omega$ so that $\mathcal{F}$ is the $\sigma$-field of subsets of $\Omega$ generated by $\mathcal{A}$. Let $\mu$ be a measure on $\mathcal{F}$ so the measurable space is $(\mathscr{F},\mu)$, where by definition any $B\in\mathscr{F}$ is $\mu$-measurable.
Suppose $\mu^{*}$ is an "outer-measure" defined as
$\mu^{*}(B)=\text{inf}\sum_{n}\mu(A_{n}),\hspace{20pt}B\subset\cup_{n}A_{n},\hspace{20pt}A_{n}\in\mathcal{A}$.
where the infimum extends over all finite or countable coverings of $B$ by $\mathcal{A}$-sets. In your question $\mu=m$ is Lebesgue measure and $\Omega=\mathbb{R}$ so that $\mathscr{F}=\mathscr{R}$ (the linear Borel sets).
Let the hypothesis you mention be true: for any $B_{1},B_{2}\in\mathscr{F}$ the following holds
$\mu^{*}(B_{1}\cup B_{2})=\mu^{*}(B_{1})+\mu^{*}(B_{2})\hspace{100pt}[1]$
Let $C_{1}=\cup_{i}A_{i,1}$, $A_{i,1}\in\mathcal{A}$ for all $i=1,2,...$ be such that $B_{1}\subseteq C_{1}$ and $\mu^{*}(B_{1})=\mu(C_{1})=\sum_{i}\mu(A_{i,1})$. Similarly let $C_{2}=\cup_{j}A_{j,2}$, $A_{j,2}\in\mathcal{A}$ for all $j=1,2,...$ be such that $B_{2}\subseteq C_{2}$ and $\mu^{*}(B_{2})=\mu(C_{2})=\sum_{j}\mu(A_{j,2})$. Writing $C_{1}\cup C_{2}$, $C_{1}$ and $C_{2}$ as a function of disjoint sets
\begin{align*} C_{1}\cup C_{2}&=C_{1}\backslash C_{2}\cup C_{2}\backslash C_{1}\cup C_{1}\cap C_{2}\\ C_{1}&=C_{1}\backslash C_{2}\cup C_{1}\cap C_{2}\\ C_{2}&=C_{2}\backslash C_{1}\cup C_{1}\cap C_{2} \end{align*}
we see
\begin{align*} \mu(C_{1}\cup C_{2})&=\mu(C_{1}\backslash C_{2})+ \mu(C_{2}\backslash C_{1})+\mu(C_{1}\cap C_{2})\\ \mu(C_{1})&=\mu(C_{1}\backslash C_{2})+\mu(C_{1}\cap C_{2})\hspace{100pt}[2]\\ \mu(C_{2})&=\mu(C_{2}\backslash C_{1})+\mu(C_{1}\cap C_{2}) \end{align*}
Using [1] and [2] we get
\begin{align*} \mu^{*}(B_{1}\cup B_{2})&=\mu^{*}(B_{1})+\mu^{*}(B_{2})\\ \Longleftrightarrow \mu(C_{1}\cup C_{2})&=\mu(C_{1})+\mu(C_{2})\\ \Longleftrightarrow \mu(C_{1}\backslash C_{2})+ \mu(C_{2}\backslash C_{1})+\mu(C_{1}\cap C_{2})&=\mu(C_{1}\backslash C_{2})+\mu(C_{2}\backslash C_{1})+2\mu(C_{1}\cap C_{2}) \end{align*}
which implies $\mu(C_{1}\cap C_{2})=0$. Since $C_{1},C_{2}\in \mathcal{A}$ we have $C_{1},C_{2}\in \mathscr{F}$ and so are both by definition measurable, and $B_{1}\subseteq C_{1}$, $B_{2}\subseteq C_{2}$. This proves your assertion.