Let $A_1,A_2, \cdots$ be an element of the borel sigma algebra such that $\lambda(A_n \cap A_m) = 0$, where $\lambda$ is of course the lesbesque measure. Let Bn be $A_n \cap A_1^c \cap A_2^c \cdots \cap A_{n-1}^c$.
I have shown that $\lambda(\cup_i A_i) = \lambda(\cup_i B_i)= \sum \lambda(B_i)$
Further i have shown:
$A_n \Delta B_n = An \subset \cup_{m=1}^{n}(A_n \cap A_m)$
Now the third question is to prove that:
$\lambda(\cup_i A_i) = \sum \lambda(A_i)$.
I don't know where to start and help would be much appreciated!
Kees