# Radon measures

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Problem

Find an example of a sequence (mn) of complex Radon measures on R that tends to 0 vaguely such that for some bounded measurable function g with compact support, $\int g\, dm_n \not\to 0$.

Definitions and facts

. complex measures are finite
. Radon measures are only defined on Borel sets
. if m is a Radon measure and E is Borel, m(E) is the infimum of the m(U), where U are open sets containing E
. if m is Radon and U is open, m(U) is the supremum of the m(K), where K are compact sets contained in U
. if m is a complex measure, there exists a positive measure m' and an m'-measurable function f such that $m(E) = \int _E f\, dm'$. We express this relation by dm = fdm'.
. if m, m', and f are as above, then the total variation of m, denoted |m| is defined by the relation d|m| = |f|dm' (this is well-defined, i.e. if m'' is another positive measure, g an m''-measurable function, such that dm = gdm'', then |f|dm' = |g|dm'')
. the norm of a complex measure m, ||m|| is defined to be $\int d|m|$
. (mn) tends to 0 vaguely iff for every continuous function f from R to C that vanishes at infinity, $\int f\, dm_n \to 0$
. if (mn) is a sequence of complex radon measures on R and Fn(x) = mn({y in R : y < x}), then if ||mn|| are uniformly bounded and the Fn converge pointwise to 0, then (mn) tends to 0 vaguely
. (there are lots theorems in this section, but none that immediately strike me as relevant)

Attempts

It seems to me that if we can find a sequence of measures and a corresponding bounded measurable function with compact support, we can do so with the restriction that this measurable function g have compact support contained in [0,1]. I've tried a few things but they haven't worked.

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## Answers and Replies

Dick
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How about thinking of delta function type measures? Take m_n to be delta(0)-delta(1/n). Then it goes to zero for any continuous function, but if we take g(x)=1 at x=0, 0 otherwise then the integral of g is 1 for any m. Does that work?

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Yes, very nice! Thanks.