Correctly defined measure I need to show that the measure is unique (correct definition).
Prove that the function $\lambda: \sigma(\mathcal{A}\cap V) \rightarrow [0,1]$ such that $\lambda[(A\cap V)\cup(B\cap V^c)]:=\mu(A)$ for $A,B\in\mathcal{A}$ and $V\subset \Omega$ is a correctly defined measure. We also know that $\mu^*(V)=1$
I guess I should take sets $A_{1}, A_{2}, B_{1}, B_{2}$ and show that whether I take $A_{1},B_{1}$ or $A_{2},B_{2}$ I still get the same measure. But I have no idea how to even start that.

Maybe it will help somebody in the future, but I think I have the solution.
We want to show that: $(A_1\cap V)\cup(B_1\cap V^c)=(A_2\cap V)\cup(B_2\cap V^c) \Rightarrow \mu(A_1)=\mu(A_2)$. In order to do that, we write:$$((A_1\cap V)\cup(B_1\cap V^c))\cap V=((A_2\cap V)\cup(B_2\cap V^c)) \cap V $$
Then: 
$$A_1\cap V=A_2\cap V $$
$$(A_1\vartriangle A_2)\cap H =\emptyset \Rightarrow (A_1\vartriangle A_2) \subset V$$
$$\nu ^*(V)\leq1-\mu(A_1\vartriangle A_2)$$
And this will be true as long as $\mu(A_1\vartriangle A_2)=0$. So $\mu(A_1)=\mu(A_2)$. So the definition is correct.
