Course in Probability Theory

384 1 SUPPLEMENT : MEASURE AND INTEGRALand consequently by property (c) :g*(Z) = g * (ZA U ZA`) < g*(ZA) + g*(ZA`) < g* (Z) .Thus (3) is satisfied and we have proved that (i) and (ii) are equivalent .Next, let A E T * and g* (A) = 0 . Then we have by the first assertion inTheorem 4 that there exists B E 3;7 such that A C B and g* (A) = g (B) . ThusA satisfies (iii) . Conversely, if A satisfies (iii), then by property (b) of outermeasure : g* (A) < g* (B) = g(B) = 0, and so (i) is true .As consequence, any subset of a (ZT*, g*)-null set is a (*, g*)-null set .This is the first assertion in the next theorem .Theorem 6 . The measure space (Q, *, g*) is complete . Let (Q, 6-7 , v) be acomplete measure space ; §' D A and v = g on A . If g is a-finite on A then;J' D3 * and v=g* on :'* .PROOF . Let A E *, then by Theorem 4 there exists B E O and C Esuch that(8) C C A C B ; g(C) = g*(A) = A(B) .Since v =,u on z o , we have by Theorem 3, v = g on 3T . Hence by (8) wehave v(B - C) = 0 . Since A - C C B - C and B - C E ;6', and (Q, ;6', v) iscomplete, we have A - C E -6- and so A = C U (A - C) E .Moreover, since C, A, and B belong to , it follows from (8) thatA(C) = v(C) < v(A) < v(B) = g(B)and consequently by (8) again v(A) = g(A) . The theorem is proved .To summarize the gist of Theorems 4 and 6, if the measure g on the field~To is a-finite on A, then (3;7 , g) is its unique extension to ~ , and ( *, g* )is its minimal complete extension . Here one is tempted to change the notationgtogoonJ%!We will complete the picture by showing how to obtain (J~ *, g*) from( ~?T , g), reversing the order of previous construction . Given the measure space(Q, ZT, g), let us denote by the collection of subsets of 2 as follows : A E Ciff there exists B E I _(:T, g) such that A C B . Clearly has the "hereditary"property : if A belongs to C, then all subsets of A belong to ~ ; is also closedunder countable union . Next, we define the collection(9) 3 ={ACQIA =B-C where BE`,CEC} .where the symbol "-" denotes strict difference of sets, namely B - C = BC`where C C B . Finally we define a function µ on :-F as follows, for the A