Let it be $I$ a finite interval and let it be $\lambda^*$ the lebesgue outer measure. Proof that $B$$\subset$$I$ is $\lambda^*$-measurable if and only if $\lambda^*(I)=\lambda^*(I$$\cap$$B)+\lambda^*(I\cap(\mathbb{R}\smallsetminus$$B))$.
I could proof that if $B$ is $\lambda^*$-measurable then $\lambda^*(I)=\lambda^*(I$$\cap$$B)+\lambda^*(I\cap(\mathbb{R}\smallsetminus$$B))$ but i don't have any idea of how to proof the other implication. If someone can give me some hints i will be really thankful.
HINT:
For every $B\subset \mathbb{R}$ we have $$\lambda^{\star}(B) = \inf_{A\supset B, A \text{measurable}}\lambda(A)$$
For $B\subset I$ we have $$\lambda(I) - \lambda^{\star}(I\backslash B) =\sup_{A\subset B, A \text{measurable}}\lambda(A)$$
For $B\subset I$, if $$\sup_{A\subset B, A \text{measurable}}\lambda(A)=\inf_{A\supset B, A \text{measurable}}\lambda(A)$$ then $B$ is measurable.