Posted by J. Holmer on December 13, 2007, 9:51 am, in reply to "Re: measure of a disjoint union of subsets of R^d"
Yes, all that you say is correct. Later on in the development of measurable sets, we find that if E_1 and E_2 are disjoint and measurable, then m(E_1 \cup E_2) = m(E_1)+m(E_2). But in order to prove this, we need the preliminary version you discuss in your post. It is fairly hard to give an example of disjoint nonmeasurable sets E_1, E_2 such that m(E_1\cup E_2) \neq m(E_1)+m(E_2), although recall we did cover this too.
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