Course in Probability Theory

3 .3 INDEPENDENCE 1 61in which there exists an r .v . X„ with µ n as its p .m . Indeed this is possible if wetake (S2„ , ?l„ , ;J/;,) to be (J/? 1 , J/,3 1 , µ„) and X„ to be the identical function ofthe sample point x in now to be written as w, (cf. Exercise 3 of Sec . 3 .1) .Now define the infinite product spaceS200= X Qnn=1on the collection of all "points" co = {c ) l, cot, . . . , cv n , . . .}, where for each n,co,, is a point of S2„ . A subset E of Q will be called a "finite-product set" iffit is of the form00(11) E = X Fn,n=1where each F n E 9„ and all but a finite number of these F n 's are equal to thecorresponding St n 's. Thus cv E E if and only if con E F n , n > 1, but this isactually a restriction only for a finite number of values of n . Let the collectionof subsets of 0, each of which is the union of a finite number of disjoint finiteproductsets, be A . It is easy to see that the collection A is closed with respectto complementation and pairwise intersection, hence it is a field . We shall takethe z5T in the theorem to be the B .F. generated by A . This ~ is called theproduct B . F. of the sequence {fin , n > 1) and denoted by X n __1 n .We define a set function ~~ on as follows . First, for each finite-productset such as the E given in (11) we set00(12) ~(E) _ fl -~ (F'n ),n=1where all but a finite number of the factors on the right side are equal to one .Next, if E E A andnE = U E (k) ,k=1where the E (k) 'sare disjoint finite-product sets, we put11(13) °/'(E (k)) .k=1If a given set E in A~ has two representations of the form above, then it isnot difficult to see though it is tedious to explain (try drawing some pictures!)that the two definitions of , ~"(E) agree . Hence the set function J/' is uniquely