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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.

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  • $\begingroup$ there is an additional hypothesis in exercise 16, that the functional be continuous. $\endgroup$ – user363464 May 30 '17 at 13:11
  • $\begingroup$ any positive linear functional is automatically continuous $\endgroup$ – Binjiu May 30 '17 at 13:12
  • $\begingroup$ ??, what is "bounded"? $\endgroup$ – Binjiu May 30 '17 at 13:16
  • $\begingroup$ why do u confirm "no"? $\endgroup$ – Binjiu May 30 '17 at 13:24
  • $\begingroup$ The norm of a linear functional $T$ is $$\lVert T\rVert = \sup\{\lvert Tx\rvert~:~\lVert x\rVert\leq 1\}.$$ A linear functional is bounded if its norm is finite. $\endgroup$ – user363464 May 30 '17 at 13:25
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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.

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    $\begingroup$ In my understanding, $T:C_{c}\left(\Omega\right)\rightarrow\mathbb{R}$ is called ``continuous'' if for any compact subset $K,$ there is $C_{K}>0$ such that $\left|\left\langle T,f\right\rangle \right|\leq C_{k}\left|f\right|_{L^{\infty}\left(\Omega\right)};$ $\forall f\in C_{c}\left(\Omega\right)$ and $supp\left(f\right)\subset K.$ Thank for your answer. It's now clear that $\mu\left(\Omega\right)$ is not necessarily finite in the sense that $T$ is continuous as this kind. $\endgroup$ – Binjiu May 30 '17 at 14:10
  • $\begingroup$ However, Tao means $T$ is continuous in a different sense, that is $\left|\left\langle T,f\right\rangle \right|\leq C_{k}\left|f\right|_{L^{\infty}\left(\Omega\right)};$ $\forall f\in C_{c}\left(\Omega\right).$ Am i right? $\endgroup$ – Binjiu May 30 '17 at 14:10
  • $\begingroup$ @Binjiu Yes. To put it in a different way, a measure is finite if and only if the constant $C_K$ can be chosen the same for ALL compact sets. $\endgroup$ – N. S. May 30 '17 at 14:15
  • $\begingroup$ @Binjiu Topologically the difference is that the space of finite measures is the dual of $(C_c(\mathbb R), \| . \|_\infty)$. Now this space is dense in $(C_0(\mathbb R), \| . \|_\infty)$, and hence they have the same dual. The space of Radon measures is the dual of $C_c(\mathbb R)$ with respect to the so called inductive topology: For each compact set $K$ the space $$C_c(\mathbb R:K):= \{ f \in C_c(\mathbb R:K) | \supp(f) \subset K \}$$ is a Banach subspace of $(C_0(\mathbb R), \| . \|_\infty)$, and now we can take the inductive limit by compact $K$. $\endgroup$ – N. S. May 30 '17 at 14:18
  • $\begingroup$ @Binju Now, it is easy to see that with respect to the inductive topology, a functional is continuous if and only if for any compact subset $K$, there exists a $C_K>0$ such that $$\left| \langle T,f \rangle \right| \leq C_{K} \| f \|_{\infty}; \forall f \in C_c\left(\mathbb R \right); supp(f) \subset K$$. So 'continuous' really means continuous with respect to the right (i.e. inductive topology). Last but not least, there are various versions of Riesz Theorem, which apply to these different situations (also positive vs signed vs complex). The equivalence between them.... $\endgroup$ – N. S. May 30 '17 at 14:23

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