need some help with the following problem in Measure Theory (couldn't find this on the forum)
Q. Prove that if A1 and A2 are measurable then
$$\lambda(A1 \bigcup A2) + \lambda(A1 \bigcap A2) = \lambda(A1) + \lambda(A2) $$
need some help with the following problem in Measure Theory (couldn't find this on the forum)
Q. Prove that if A1 and A2 are measurable then
$$\lambda(A1 \bigcup A2) + \lambda(A1 \bigcap A2) = \lambda(A1) + \lambda(A2) $$
Hint. Let $C = A_1 \cap A_2, B_1 = A_1 \setminus C, B_2 = A_2 \setminus C$. Then $C, B_1, B_2$ are disjoint measurable sets. These sets have nice addition properties you can use.
$\lambda(A_1) + \lambda(A_2) = \lambda( A_1 \cap A_2 ) + [ \lambda(A_1 - A_2) + \lambda(A_2 - A_1) + \lambda (A_1 \cap A_2)] = \lambda (A_1 \cap A_2) + \lambda(A_1 \cup A_2) $
You can draw a vent diagram if you have a hard time to see this.