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!


  • $\begingroup$ Use that $A_n\setminus B_n$ is contained in a finite union of measure zero sets $A_n\cap A_k$, $k=1,2,...,n-1$. Then compare the measures of $A_n$ and $B_n$. $\endgroup$ – Pp.. Feb 6 '15 at 17:22
  • $\begingroup$ I don't see how $A_1^c \cap \cdots A_{n-1}^c$ is contained in that set to be honest... Thanks for the hint though Kees $\endgroup$ – Kees Til Feb 6 '15 at 18:01
  • $\begingroup$ o crap never mind i did the $B_n\A_n$ $\endgroup$ – Kees Til Feb 6 '15 at 18:02
  • $\begingroup$ i got eventually this, i don't know for sure if it's right. $\lambda (A_n \ B_n) = \lambda (\cup_{i<n}A_n \cap A_i) \leq \sum \lambda(An \cup A_i) = 0.$ because $B_n \subset A_n$ we see $\lambda B_n = \lambda A_n$ $\endgroup$ – Kees Til Feb 6 '15 at 18:32
  • 1
    $\begingroup$ Yes, that's good. $\endgroup$ – Pp.. Feb 6 '15 at 19:55

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