Radon measure

(redirected from Outer regular)

Radon measure

[′rā‚dän ‚mezh·ər]
(mathematics)
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
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(i) v is inner regular by closed sets and outer regular by open sets;
Now we can define a v: F [right arrow] E, v(B) = [??](A), A being any element in A with B = A [intersection] X; it is a trivial verification that v is well-defined, is finitely additive and it is inner regular by closed sets in X and outer regular by open sets in X (this means for a p [member of] P, F [member of] F and c > 0, there is, in X, a closed set C and an open set V, C [subset] F and V [contains] F, such that for any [B.sub.1] [member of] F, [B.sub.1] [subset] F\C and any [B.sub.2] [member of] F, [B.sub.2] [subset] V\F, we have [[parallel]v([B.sub.i])[parallel].sub.p] < c for i = 1,2).
Let v: F [right arrow] E be a finitely additive regular (inner regular by closed sets in X and outer regular by open sets in X) measure, having relatively weakly compact range, such that [integral] fdv = 0, [for all]f [member of] [C.sub.b](X).
Then it has a unique extension to a countably additive Borel measure [mu]: B(X) [right arrow] E which is inner regular by closed sets and outer regular by open sets.
Take any B [member of] B(X) and select any [??] [member of] A such that [??][intersection] X = B; define [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]; it is a trivial verification that [mu] is well-defined, is countably additive and it is inner regular by closed sets in X and outer regular by open sets in X.