By Satoru Igari

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Thus 0 E B(~d), which implies that u(O(~d)) C B(~d). On the other hand, K(~d) C u( O(~d)), since the complement of a compact set is open. Thus B(~d) c u( O(~d)). Combining this with the reverse inclusion, we get B(~d) = u( O(~d)). Let 0 be a nonempty open set in ~d. Then 0 E M (~d), since 0 is a countable union of mutually disjoint right half-open intervals LJ Ij and Ij E M (~d). Thus O(~d) c M(~d). Thus B(~d) cM (~d). Let N(~d) be the collection of all null sets. 3, N(~d) c M(~d). 6. M(~d) is the smallest u-field which includes B(~d) and N(~d), that is, M = u(B UN).

Since {( s, t) :

a} is an open set in ~ 2 , it is a countable union of mutually disjoint right half-open intervals [a;, b;) x [c;, d;). Thus, 00 {x E ~d:

a}= LJ{x : (f(x),g(x)) E [a;,b;) x [c;,d;)} j=l 00 = LJ{x: a;$ f(x) < b;}n{x: c; $ g(x) < d;}. j=l The last sum is measurable since each summand is measurable. 1. (i) The set of all complex valued measurable functions is a vector space; that is, if f and g are measurable functions and a, b E C, then af + bg is measurable.

JRd) (j = 1, ... 3) m(E):::; L m(E;). j=l c u;=l E;, then 2. 26 LEBESGUE MEASURE ON EUCLIDEAN SPACES PROOF. Put F; =En E;. We can write E as follows: E = En k k j=i j=i LJ E; = LJ F; =Fi U (F2 - Fi) U (Fa - Fi - F2) U · · · U (Fk - Fi - · · · - Fk-i). The right-hand side is a union of mutually disjoint sets in n(JRd). 2) we have + m(F2 - m(E) =m(Fi) + .. · + m(Fk - Fi) + m(Fa - Fi - F2) Fi - .. · - Fk_i). To estimate the right-hand side use the inequalities m(Fi):::; m(Ei), m(F2 - Fi):::; m(E2), ...