1
$\begingroup$

This is the question, where $|*|$ denotes Lebesgue measure.

Let ${E_j}$ be a sequence of Lebesgue measurable sets in $\Bbb{R}^n$ st. $|E_j⋂E_i |=0$ for $j≠i$ (i.e. they are pairwise disjoint a.e.), show that $|⋃E_j |=∑|E_j |$.

My solution. I am not confident because I recall the textbook took a lot hustle and bustle to prove the disjoint case (in this question it is disjoint a.e.). But I am not able to find anything wrong in my proof. Hope someone can help me verify it. Thank you!

Denote $E_{ij}=E_j⋂E_i$, then $|E_{ij}|= 0$ and $⋃_j (E_j\backslash (⋃_i E_{ji}))⊆⋃E_j$. Also note $⋃_i E_{ji}$ is a countable union of null sets and thus still null, and $|E_j\backslash (⋃_i E_{ji})|=|E_j |$ because removal of a null subset does not change measure.

Thus $|⋃E_j |≥|⋃_j (E_j\backslash (⋃_i E_{ji}))|≥∑_j|E_j\backslash (⋃_i E_{ji})|=∑_j|E_j |$. The opposite inequality is obvious, so the claim holds.

$\endgroup$
  • $\begingroup$ $E_j\backslash (⋃_{i\neq j} E_{ji}))$ or else it will be empty set. $\endgroup$ – Xiao Jul 25 '15 at 4:21
1
$\begingroup$

Your proof looks fine, other than the issue noticed by Xiao that you should exclude $E_{jj}$ from the union of sets being subtracted.

Note that the inequality $$|⋃_j (E_j\backslash (⋃_i E_{ji}))|≥∑_j|E_j\backslash (⋃_i E_{ji})|$$ is really an equality, since there is a disjoint union on the left side. So here you are using the "disjoint" case of countable additivity, which as you saw was more difficult to prove. So it's not surprising that your proof seemed much easier: you are taking advantage of the hard part already being done.

Did you perhaps think you were only using the countable subadditivity in this line? That inequality goes the other direction. You use countable subadditivity in your last sentence; it is what justifies the "obvious" opposite inequality (you may want to reconsider whether the word "obvious" is warranted here).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.