# Riesz representation theorem in measure theory

Let $\Omega\subseteq\mathbb{R}^{n}$ be an open set and let $C_{c}\left(\Omega\rightarrow\mathbb{R}\right)$ be the space of continuous, compactly supported functions on $\Omega.$

We now consider two different topologies on $C_{c}\left(\Omega\rightarrow\mathbb{R}\right):$ $$T_{1}:=\left[C_{c}\left(\Omega\rightarrow\mathbb{R}\right);\left|\cdot\right|_{L^{\infty}\left(\Omega\right)}\right]\quad and\quad T_{2}:=\left[C_{c}\left(\Omega\rightarrow\mathbb{R}\right);\textrm{the inductive topology}\right].$$ We have $T_{1}$ a normed space but $T_{2}$ just a topological vector space in general. We now consider "continuous" (w.r.t these topologies) linear functionals on $T_{1}$ and $T_{2}:$ $$F_{1}:T_{1}\rightarrow\mathbb{\mathbb{R}}\quad and\quad F_{2}:T_{2}\rightarrow\mathbb{R}.$$ (By definition, the statement of continuity is equivalent to: there exists $C>0$ such that $\left|\left\langle F_{1},f\right\rangle \right|\leq C\left|f\right|_{L^{\infty}\left(\Omega\right)}$; $\forall f\in C_{c}\left(\Omega\rightarrow\mathbb{R}\right)$; and for any compact subset $K\subseteq\Omega,$ there exists $C_{K}>0$ such that $\left|\left\langle F_{2},f\right\rangle \right|\leq C_{K}\left|f\right|_{L^{\infty}\left(\Omega\right)};$ $\forall f\in C_{c}\left(\Omega\rightarrow\mathbb{R}\right),suppf\subseteq K.$)

From Exercise 16, [https://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/#rrt] there exists a unique, finite (signed) Radon measure $\mu$ on $\Omega$ such that $$\left\langle F_{1},f\right\rangle =\intop_{\Omega}fd\mu;\forall f\in C_{c}\left(\Omega\rightarrow\mathbb{R}\right).$$

My question is: is it true that $$\left\langle F_{2},f\right\rangle =\intop_{\Omega}fd\eta;\forall f\in C_{c}\left(\Omega\rightarrow\mathbb{R}\right);$$ for some (unique, but may not be finite) signed Radon measure $\eta$ on $\Omega?$

Attempt: I think it's true. First, in the case of "positive linear functional", we already know it is the case [Theorem 8, https://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/#rrt]. Therefore, we firstly prove that: any continuous linear functional $F_{2}$ can be represented as $F_{2}=H-G,$ with $H:T_{2}\rightarrow\mathbb{R}$ and $G:T_{2}\rightarrow\mathbb{R}$ positive linear functionals. But what is the point of using the inductive topology in this stage?

• With inductive topology do you mean the topology where the pointwise evaluations are to be continuous? From some of what you have written it seems you mean the topology given by $$\lim_{K\subset\Omega} (C(K),\|\cdot\|_{K,\infty})$$ where $K$ are compact subsets, which I believe is the same as $C_c(\Omega)$. – s.harp May 31 '17 at 13:43
• For the definition of the inductive limit topology: en.wikipedia.org/wiki/Distribution_(mathematics), it's like the def. of the topology of test functions in the distribution theory, but here we consider $C_{c}\left(\Omega\right)$ instead of $C_{c}^{\infty}\left(\Omega\right)$. – Binjiu Jun 1 '17 at 1:37
• What is your definition of $C_{c}(\Omega)$? – Binjiu Jul 5 '17 at 6:42
• $C_c(\Omega)$ is the space of compactly supported continuous functions on $\Omega$ together with sup norm $\|\phi\|=\sup_{x\in\Omega}|\phi(x)|$. – s.harp Jul 5 '17 at 10:57
• Right now I cannot remember my argument that the two topologies are the same, and am unsure if it is true. However $T_2$ is finer than $T_1$, so any functional that is continuous on $T_2$ must be continuous on $T_1$. – s.harp Jul 5 '17 at 11:18

You can localize Riesz Theorem on the first space to obtain it in the second one. Choose some continuous functions $\phi_i \colon \mathbb R^n \to \mathbb R$ such that $\phi_i=1$ on $B_i(0)$ (the ball with radius $i$) and $\phi_i=0$ outside $B_{i+1}(0)$. Then define for $f\in T_2$ $$F_2^i(f) := F_2(f \cdot \phi_i).$$ Notice that when $f\in T_2$ also $f\cdot \phi_i\in T_2$. Notice also that $f\cdot \phi_i \to f$ in $T_2$. So, $F_2^i(f) \to F_2(f)$ for each $f\in T_2$.
Moreover each $F_2^i$ is also continuous on $T_1$ hence there exists a finite Radon measure $\mu^i$ such that $F^i_2(f) = \int f \, d\mu^i$.
Now we have $$F_2(f) = \lim_i F_2^i(f) = \lim_i \int f \, d \mu^i = \int f \, d \mu$$ where in the last equation we have introduced the set function $\mu(E) := \lim_i \mu_i(E)$. This set function coincides with $\mu_i$ in $B_i(0)$ hence is actually a measure itself (* this point should be clarified, but somewhat depends on how non finite signed measures have been defined *)