Radon measure, Riesz's theorem Let $\Omega\subseteq\mathbb{R}^{n}$ be an open set and 
$$
T:C_{c}\left(\Omega\right)\rightarrow\mathbb{R}
$$
 be a positive linear functional, where $C_{c}\left(\Omega\right):=\left\{ f:\Omega\rightarrow\mathbb{R}:f\textrm{ is continous, compactly support}\right\} .$
By Riesz's theorem, there is a (unique) positive Radon measure $\mu$
such that 
$$
\left\langle T,f\right\rangle =\intop_{\Omega}fd\mu;\forall f\in C_{c}\left(\Omega\right).
$$
My question is: is it always true that $\mu\left(\Omega\right)<+\infty$? 
In Exercise 16, https://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/, the author claimed that it's true. Am i wrong?
Thank.
 A: No.
Take $\Omega = {\mathbb R}^n$ or any open unbounded set with infinite Lebesgue measure,  and let $T f = \int f(x) dx$. 
Even if $\Omega$ is bounded, the answer is still negative. Here's a simple example. Let $(a_n)$ be a sequence in $\Omega$ converging to a point in $\partial \Omega$, and let 
$T f =\sum_n f(a_n) $. 
If $f$ is compactly supported, then the sum is finite. If, in addition, $f$ is nonnegative, then the sum is nonnegative. 
Also, by definition, $Tf$ is the integral of $f$ with respect to an infinite measure $\sum_n \delta_{a_n}$. 
Finally, for Tao's notes. I did not read thoroughly, but I seriously doubt that there is such a claim. If you're referring to Theorem 8, then note that a Radon measure is not necessarily finite (one of the first examples is Lebesgue measure). If you're referring to Exercise 16, note that the assumption $I \in C_c(\Omega)^*$ is somewhat stronger, and excludes the two examples above, the reason being $C_c(\Omega)^*=C_0(\Omega)^*$ ($C_0(\Omega)$ is the  space of continuous functions ``vanishing at $\infty$", equipped with the $\sup$-norm. This  is a Banach space), and the functionals defined in the two examples above are not bounded on $C_0(\Omega)$. Recall that  $C_c({\Omega})$ is NOT a Banach space: its closure is $C_0({\mathbb \Omega})$. Perhaps this is the source of confusion.
