# Construction of measure in Riesz Repr. Thm. (Rudin)

In the proof of Riesz Representation Theorem in Rudin; Real and Complex Analysis (theorem 2.14, page 41/42), regarding construction of the measure, Rudin goes as follows.

For every open set V in X define: \begin{equation} \mu (V) = \sup \{ \Lambda f:f \prec V\} \label{1} \end{equation}

If ${V_1} \subset {V_2}$ also $\mu ({V_1}) \le \mu ({V_2})$

And now the tricky part, my question is, why is the following a consistent definition? and why for every $E$? It seems reasonable since X is locally compact Hausdorff according to the assumptions in the theorem, but I am unable to come up with a rigorous proof. The consistent definition: $$\mu (E) = inf\{ \mu (V):E \subset V,V\,open\}$$ if $E$ is an open set, and $\mu(E)$ is consistent to the before mentioned definition for every $E\subset{X}$.

Note: $$f \prec V$$ Is notation for that $V$ is open, $f \in {C_c}(X)$, i.e. $f$ is continuous with compact support, $0 \le f \le 1$ and the support lies in $V$.

Also $\Lambda$ is a positive linear functional.

Thank you very much in advance!

Let me change the notation slightly. For every open set $V \subset X$, define $$\mu_{\rm for \ open}(V) = \sup \{ \Lambda f : f \prec V \}.$$ Next, for every set $E \subset X$ (open or otherwise), define $$\mu_{\rm general}(E) = \inf \{ \mu_{\rm for \ open}(V) : E \subset V, V {\rm \ open} \}.$$
When Rudin says that the two definitions are consistent for open sets, he means that $$\mu_{\rm general}(E) = \mu_{\rm for \ open}(E)$$ in the special case where $E$ is itself an open set.
So why is this true? Let's look at the definition of $\mu_{\rm general}(E)$, under the assumption that $E$ itself is open. Observe that, for any open $V$ such that $E \subset V$, we have $$\mu_{\rm for \ open}(E) = \sup \{ \Lambda f : f \prec E \} \leq \sup\{\Lambda f : f \prec V \}\leq \mu_{\rm for \ open}(V).$$ Hence $$\mu_{\rm for \ open}(E)\leq \inf \{ \mu_{\rm for \ open}(V) : E \subset V , V {\rm \ open}\}$$
Furthermore, by considering $V = E$, (which certainly is an open set containing $E$, since $E$ is assumed to be open), we find that $$\inf\{\mu_{\rm for \ open}(V) : E \subset V, V {\rm \ open} \} \leq \mu_{\rm for \ open}(E)$$
Combining these two inequalities gives $\mu_{\rm general}(E) = \mu_{\rm for \ open}(E)$ is the case where $E$ is an open set, which is the desired statement of consistency.