Exercise 4 of Chapter 6 in Rudin's Functional Analysis states that every "positive" distribution $\Lambda\in D^{'}(\Omega)$, i.e, $\Lambda\psi\geq 0$ whenever $\psi\in D(\Omega)$, is a positive measure in $\Omega$, where $\Omega\subseteq\mathbb{R}^d$ open and $ D^{'}(\Omega)$ is the space of test functions on it.
My question is that does problem directly follow from the Riesz Representation theorem that says every positive linear functional on the space of compactly supported smooth functions on a locally compact Hausdorff space is a positve Radon measure?
Best Answer
It's actually easier than it might seem.
Just notice that if $T$ is a positive distribution then $(T,\phi)\geq(T,\psi)$ for $\phi\geq \psi$ (since $\phi-\psi\geq 0$ implies $(T,\phi-\psi)\geq 0$). Therefore, for each $K\subset\subset D$, $$\sup_{\phi\in D_K,\,||\phi||_{C^0}\leq 1}(T,\phi)\leq(T,\eta_K) <\infty,$$ where $\eta_K\in D$ is some positive cut-off on $K$. This means that $T$ has order $0$. Thus it can be extended to $C_c(\Omega)$ and so, by Riesz's Theorem, it is a measure.