1.3. The Lebesgue integral 43 holds identically on Rd. Then one has c1m(E1) + . . . + ckm(Ek) = c1m(E1) + . . . + ck m(Ek ). Proof. Again, we use a Venn diagram argument. The k+k sets E1,...,Ek, E1,...,Ek partition Rd into 2k+k disjoint sets, each of which is an inter- section of some of the E1,...,Ek,E1,...,Ek and their complements. We throw away any sets that are empty, leaving us with a partition of Rd into m non-empty disjoint sets A1,...,Am for some 0 ≤ m ≤ 2k+k . As the E1,...,Ek,E1,...,Ek are Lebesgue measurable, the A1,...,Am are too. By construction, each of the E1,...,Ek,E1,...,Ek arise as unions of some of the A1,...,Am, thus we can write Ei = j∈Ji Aj and Ei = j ∈J i Aj for all i = 1,...,k and i = 1,...,k , and some subsets Ji,Ji ⊂ {1,...,m}. By finite additivity of Lebesgue measure, we thus have m(Ei) = j∈Ji m(Aj) and m(Ei ) = j∈J i m(Aj) Thus, our objective is now to show that (1.10) k i=1 ci j∈Ji m(Aj) = k i =1 ci j∈J i m(Aj). To obtain this, we fix 1 ≤ j ≤ m and evaluate (1.9) at a point x in the non-empty set Aj. At such a point, 1Ei (x) is equal to 1Ji (j), and similarly 1E i is equal to 1J i (j). From (1.9) we conclude that k i=1 ci1Ji (j) = k i =1 ci 1J i (j). Multiplying this by m(Aj) and then summing over all j = 1,...,m we obtain (1.10). We now make some important definitions that we will use repeatedly in this text:

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2011 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.