Suppose that $\mu$ is a Radon measure on a locally compact Hausdorff space $X$ and $\phi\in C(X,(0,\infty))$. Let $\nu(E)=\int_{E}\phi\;d\mu$ and let $\nu'$ be the Radon measure associated to the functional $f\mapsto\int f\phi\;d\mu$ on $C_{c}(X)$ (so $\int f\phi\;d\mu=\int f\;d\nu'$ for all $f\in C_{c}(X)$). Show that $\nu(U)=\nu'(U)$ for any open set $U$.

I have shown that $\nu(U)\geq\nu'(U)$ by noting that $\phi\chi_{U}$ is nonnegative and lower semicontinuous. How should I get the reverse inequality?

Next I want to show that $\nu$ is outer regular on all Borel sets by observing that the open sets $V_{k}=\{x:2^{k}<\phi(x)<2^{k+2}\}$ ($k\in\mathbb{Z}$) cover $X$.

Finally I want to show that $\nu=\nu'$.

  • $\begingroup$ I now have the reverse inequality for the first part. The last part follows from the first two. I am still having problems with the second part. $\endgroup$ – cyc Apr 14 '13 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.