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The problem is to show

Suppose that $\mu$ is a Radon measure on $X$, If $\phi \in L^1(\mu)$ and $\phi \geq 0$,then prove that $\nu(E)=\int_E \phi d\mu$ is a Radon measure.

My attempt:

We first show the following equality $$ (*) \int_X f d\nu = \int_X f \phi d\mu, \enspace \forall f \in C_c(X). $$ By assumption, $(*)$ clearly holds for the characteristic functions of Borel sets, as we can see from equation below. $$ \int_X \chi_E d\nu = \nu(E) =\int_E \phi d\mu= \int_X \chi_E \phi d\mu. $$ It then follows from monotone convergence theorem and the fact that any non-negative measurable function is an increasing limit of a sequence of simple functions that $(*)$ actually holds for all non-negative measurable $f$. Since we can decompose any $f \in C_c(X)$ into positive and negative part, $(*)$ also holds for such $f$.

We now consider the linear functional defined by $$ I(f) := \int_X f d\nu, \enspace (f \in C_c(X)). $$ Since $\phi \geq 0$, $I$ is a positive linear functional. Hence, by the uniqueness given by Riesz Representation theorem, we conclude that $\nu$ is Radon.

The text has given a hint to use Corollay 3.6. However, I'd like to know whether my own proof is also correct or not. Thanks.

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1 Answer 1

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No. What you have proved here is that there exists a Radon measure $\tilde{\nu}$ such that

$\displaystyle \int_X f \, d \tilde{\nu} = \int_X f \, d \nu$

for all $f \in C_c(X)$, but not necessarily that $\nu$ is Radon.

The uniqueness in Riesz's representation theorem only comes from the fact that we require some regularity assumptions on the measures (i.e. that they are Radon). Hence what you need to show to solve the problem is that $\nu$ satisfy the regularity assumptions given by being Radon.

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