1.7. Outer measures, pre-measures, product measures 157 and also adopting the convention that the empty interval has zero F - volume, |∅|F = 0. Note that F−(+∞) could equal +∞ and F+(−∞) could equal −∞, but in all circumstances the F -volume |I|F is well defined and takes values in [0, +∞], after adopting the obvious conventions to evaluate expressions such as +∞ (−∞). A somewhat tedious case check (Exercise!) gives the additivity property |I J|F = |I|F + |J|F whenever I, J are disjoint intervals that share a common endpoint. As a corollary, we see that if an interval I is partitioned into finitely many disjoint sub-intervals I1,...,Ik, we have |I| = |I1| + . . . + |Ik|. Let B0 be the Boolean algebra generated by the (possibly infinite) inter- vals, then B0 consists of those sets that can be expressed as a finite union of intervals. (This is slightly larger than the elementary algebra, as it allows for half-infinite intervals such as [0, +∞), whereas the elementary algebra does not.) We can define a measure μ0 on this algebra by declaring μ0(E) = |I1|F + . . . + |Ik|F whenever E = I1 . . . Ik is the disjoint union of finitely many intervals. One can check (Exercise!) that this measure is well defined (in the sense that it gives a unique value to μ0(E) for each E B0) and is finitely additive. We now claim that μ0 is a pre-measure thus we suppose that E = B0 is the disjoint union of countably many sets E1,E2,... B0, and wish to show that μ0(E) = n=1 μ0(En). By splitting up E into intervals and then intersecting each of the En with these intervals and using finite additivity, we may assume that E is a single interval. By splitting up the En into their component intervals and using finite additivity, we may assume that the En are also individual intervals. By subadditivity, it suffices to show that μ0(E) n=1 μ0(En). By the definition of μ0(E), one can check that (1.35) μ0(E) = sup K⊂E μ0(K) where K ranges over all compact intervals contained in E (Exercise!). Thus, it suffices to show that μ0(K) n=1 μ0(En)
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