Suppose λ is the Lebesgue measure, $A_0$ is the $\sigma$-algebra of Lebesgue measurable sets, and E a non-measurable set with $0=λ_\ast(E)$ , $λ^\ast(E)=1$ , $\nu(E) = \gamma$.
let $ \lambda [(A ∩ E) ∪ (B ∩ (X \setminus E))] = γλ(A) + (1 − γ)λ(B)$. Where $A,B\in A_0$.
then by extension theorem this is a measure on the $\sigma$-algebra $\sigma(A_0 ∪ \{E\})$.
For me is not easy to see that this is a measure. Clearly, not hard to see that for empty set this is zero and also that this is non-negative. But I don't know how to prove that it satisfies the additivity property.
$\textbf{My attempt}$ to prove the additivity, but not sure if it is correct;
let $\cup D_i$ be the disjoint union of the sets where $D_i \in \sigma(A_0 ∪ \{E\})$ , (i.e. $D_j\cap D_i = \emptyset $). By definition we have that
$$D_i = (A_i ∩ E) ∪ (B_i ∩ E^c)$$ $$\bigcup_{i=1}^{\infty} D_i = \bigcup_{i=1}^{\infty}\left[(A_i ∩ E) ∪ (B_i ∩ E^c) \right]$$ $$\lambda(\bigcup_{i=1}^{\infty} D_i) = \lambda(\bigcup_{i=1}^{\infty}\left[\underbrace{(A_i ∩ E) ∪ (B_i ∩ E^c)}_\text{disjoint}\right])$$
since $D_j\cap D_i = \emptyset \implies [(A_j ∩ E) ∪ (B_j ∩ E^c) ] \cap [(A_i ∩ E) ∪ (B_i ∩ E^c)] = \emptyset$ \begin{equation} \lambda(\bigcup_{i=1}^{\infty} D_i) = \sum_{i=1}^{\infty}\lambda(\left[\underbrace{(A_i ∩ E) ∪ (B_i ∩ E^c)}_\text{but is it $\lambda$-measurable????}\right]) \end{equation} $$\lambda(\bigcup_{i=1}^{\infty} D_i) = \sum_{i=1}^{\infty}\lambda(D_i)$$