Prove that if $A$ and $B$ are measurable, then $\lambda(A)+\lambda(B)=\lambda(A \cup B)+\lambda(A \cap B)$

Question

Prove that if $A$ and $B$ are measurable, then $\lambda(A)+\lambda(B)=\lambda(A \cup B)+\lambda(A \cap B)$

I tried to prove it using

$A\cup B =(A\cap B^c)\cup (A\cap B)\cup(A^c\cap B)$

but failed. Give me other hints. Thank you!

Answer 1

You are on the right track. The equality you mention results in: $$\lambda(A\cup B)=\lambda(A\cap B^c)+\lambda(A\cap B)+\lambda(A^c\cap B)$$ because we are dealing with a union of disjoint sets. Adding up $\lambda(A\cap B)$ on both sides gives: $$\lambda(A\cup B)+\lambda(A\cap B)=\lambda(A\cap B^c)+2\lambda(A\cap B)+\lambda(A^c\cap B)$$ Now apply the equalities $\lambda(A)=\lambda(A\cap B)+\lambda(A\cap B^c)$ and $\lambda(B)=\lambda(A\cap B)+\lambda(A^c\cap B)$ on RHS and you are ready.

Answer 2

Hint:use the definition of mesurability to find two equations

Answer 3

You want to determine $\lambda(A\cup B)$. The property of the measure which can help you here is the $\sigma-$additivity. But in order to use it we must have disjoint sets in the argument of $\lambda$.

We can write: $A\cup B=A\(A\cap B)\cup B\(A\cap B)\cup(A\cap B)$. The RHS is an union of disjoint sets. (Make a sketch!).

Then $\lambda(A\cup B)=\lambda(A)+\lambda(B)-2\lambda(A\cap B)+\lambda(A\cap B)$.

Rearranging yields the desired result.