Rudin’s RCA: $2.14$ theorem

analysisfunctional-analysismeasure-theoryreal-analysis

There are the definitions which we need for the proof of the theorem:
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There is the Urysohn's lemma which we need for the proof:
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There is the theorem:

Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a
positive linear functional on $C_{c}(X)$. Then there exists a $\sigma$-algebra $\mathfrak{M}$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathfrak{M}$ which represents $\Lambda$ in the sense that:

  1. $\Lambda f = \int_X f d\mu \;\; \forall f \in C_c(X)$;
  2. $\mu(K) < \infty \forall K \subset X$, $K$ compact set;
  3. $\forall E \in \mathfrak{M}$, we have
    $$\mu(E) = inf \left\{ \mu(V) : E \subset V, V \text{ open} \right\}$$.
  4. The relation
    $$ \mu(E) = sup \left\{ \mu(K) : K \subset E, K \text{ compact} \right\}$$.
    Holds for every open set $E$, and for every $E \in \mathfrak{M}$ with $\mu(E) < \infty$
  5. If $E \in \mathfrak{M}$, $A \subset E$ and $\mu(E) = 0$, then $A \in \mathfrak{M}$.

There is the proof:

Let us begin by proving the uniqueness of $\mu$. if $\mu$ satisfies $(c)$ and $(d)$, it is clear that $\mu$ is determined on $\mathfrak M$ by its values on compact sets. Hence it suffices to prove that $\mu_1(K)$ $=$ $\mu_2(K)$ for all $K$, whenever $\mu_1$ and $\mu_2$ are measures for which the theorem holds. So, fix $K$ and $\epsilon$ $\gt$ $0$. By $(2)$ and $(3)$, there exists a $V$ $\supset$ $K$ with $\mu_2(V)$ $\lt$ $\mu_1(K)$ $+$ $\epsilon$; by Urysohn's lemma, there exists an $f$ so that $K$ $\prec$ $f$ $\prec$ $V$;

hence

$$ \mu_1(K) = \int_X\chi_K\,d\mu_1 \le \int_X f\,d\mu_1 = \Lambda f = \int_X f\,d\mu_2 \le \int_X \chi_V\,d\mu_2 = \mu_2(V) < \mu_2(K) + \epsilon. $$

I don't understand why is $\int_X f d\mu_1$ equal of $\int_X f d\mu_2$ ?

Any help would be appreciated.

Best Answer

$\newcommand{\d}{\,\mathrm{d}}$Rudin writes: "$\mu_1$ and $\mu_2$ are measures for which the theorem holds". In particular the assumption/assertion $(1)$ holds for both $\mu_{1,2}$. That is: $$\Lambda f=\int_Xf\d\mu_1\\\Lambda f=\int_X\d\mu_2$$For all $f\in C_c(X)$. So then $\int_X f\d\mu_1=\int_Xf\d\mu_2$ follows (by transitivity of equality)

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