TITLE: Riesz representation theorem in measure theory
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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 *)