Game Theory

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The proofs of the other implications are similar, and are left as exercises (which you should attempt). We have to show that for any interval I, j-l(I) E M. By (b) we have that for the particular case I = (a,oo). Suppose 1= (-oo,a]. Then since both E and j-l((a, (0)) are in M (we use the closure properties of M established before ). Next j-l ( 1 U(-00, b - ~ l) 00 n=1 Ur 00 1 ((-00,b- 1 ~l). 1), j-l (( -00, b - ~]) E M and the same is true for the countable union. From this we can easily deduce that Now let 1= (a, b), and j-l(( -00, b) n (a, (0)) r 1 (( -00, b)) n r 1 ((a, (0)) 57 3.

U::: 41 2. 5 Put ß = n{F: Fis a a-field containing all intervals}. We say that ß is the a-field generated by all intervals and we call the elements of ß Borel sets. It is obviously the smallest a-field containing all intervals. 1 (Borel sets) The following examples illustrate how the closure properties of the a-field ß may be used to verify that most familiar sets in IR belong to ß. (i) By construction, all intervals belong to ß, and since ß is a a-field, all open sets must belong to ß, as any open set is a countable union of (open) intervals.

This implies that the sets ZA and ZA+t are the same so their infima are equal. 6 (ii) all Eh are in M, hence by (iii) the same can be said ab out the union U~l E k· Finally, by (ii) again, the complement of this union is in M, and so the intersection n~l Ek is in M. 1 tells us that the outer measure is monotone, but since m is just the restriction of m* to M, then the same is true for m: A C B implies m(A) = m*(A) ::; m*(B) = m(B). (ii) We write B as a disjoint union B = AU (B \ A) and then by additivity of m we have m(B) = m(A) + m(B \ A).