Theorem 2.14 in Rudin's RCA: What is meant by "(a) forces (b)"?

Question

Theorem 2.14 is the Riesz Representation Theorem(about functionals on $C_c(X)$):

Theorem 2.14. Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a 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 mesaure $\mu$ on $\mathfrak M$ which represents $\Lambda$ in the sense that
(a) $\Lambda f = \int_X fd\mu$ for every $f \in C_c(X)$,
and which has the following additional properties:
(b) $\mu(K)<\infty$ for every compact set $K\subset X$.
(......the left omitted)

And a few lines later the author writes:

$$ \mu_1(K) = \int_X\chi_K\,d\mu_1 \le \int_X f\,d\mu_1 = \Lambda f = \int_X f\,d\mu_2 \le \int_X \chi_V\,d\mu_2 = \mu_2(V) < \mu_2(K) + \epsilon. $$ Incidentally, the above computation shows that (a) forces (b).

"(a) forces (b)" is the problem. How this is possible? The author has just used (b) to show the above inequality, and says that (a) implies (b) by the above computation. I need explanation about this.

Answer 1

If we know $\Lambda f = \int_X fd\mu$ for some $\mu$, then if $K$ is compact we can find some function $g$ that is $1$ on $K$ and compactly supported on $X$ and then this integral is at least $\mu(K)$ and must also be finite, as $\Lambda(f)$ is. So $\mu$ is then always be finite on compact sets. So for $(a)$ to hold at all, $(b)$ must be satisfied for the $\mu$ we want. I think that's all he's saying.

Answer 2

Let $K$ be compact there exists an open set $V\supset K$ and by Ursoyhn's lemma there exists an $f$ such that $K \prec f \prec V $ so that $0\leq f \leq 1$ and $f(X)\subset[0,1]\subset[0,\infty)$ which implies $\Lambda f\in [0,\infty)$ because $\Lambda$ is a positive linear functional so $\Lambda f$ is finite then since $\chi_K \leq f \leq \chi_V$ (by Ursoyhn's lemma) we have;

$$\begin{aligned} \mu(K)& =\int_X \chi_K d\mu \\& \leq \int_X f d\mu \\&= \Lambda f \\&< \infty \end{aligned}$$

so $\mu(K)$ is finite.