Rudin's Proof on Riesz Representation Theorem

Question

The proof is from Rudin's Real and Complex Analysis. I am having a hard time understanding part of the Riesz Representation Theorem

The Theorem states: Every open set $E$ satisfies $$\mu(E)=sup\,\{\mu(K):K\subset E,\,K\,\,\text{compact}\}\,\,\,\,\,\,(3)$$ (Note: Here $\mu$ is a positive measure, $\Lambda$ is a positive linear functional and $\mathfrak M_F$ is the class of all $E\subset X$ that satisfies $(3)$ and $\mu(E)<\infty$.)

Hence $\mathfrak M_F$ contains every open set $V$ with $\mu(V)<\infty$

The proof goes as: Let $\alpha$ be a real such that $\alpha<\mu(V)$. There exists $f\prec V$ with $\alpha<\Lambda f$. If $W$ is any open set which contains the support $K$ of $f$, then $f\prec W$, hence $\Lambda f\leq\mu(W)$. Thus $\Lambda f\leq \mu(K)$. This exhibits a compact $K\subset V$ with $\alpha<\mu(K)$, so that $(3)$ holds for $V$.

Here is what I understand along with what I don't: $\alpha<\mu(V)$ is clear since $\mu(V)$ is a positive measure. I am not sure whether there exists an $f\prec V$ with $\alpha<\Lambda f$. I think that the reasoning comes from the definition that $$\mu(V)=\sup\{\Lambda f:f\prec V\}$$ where $V$ is any open set in $X$. Here the $f\prec V$ with $\alpha<\Lambda f$ would be the supremum. However, I am not sure if that is true.

The fact that if $W$ is any open set which contains the support $K$ of $f\Rightarrow f\prec W\Rightarrow \Lambda f\leq \mu(W)$ is clear. However, I don't understand why this implies that $\Lambda f\leq\mu(K)$. Nor do I understand how $K\subset V$. Lastly, I don't think that I understand why the fact that $K\subset V$ and $\alpha<\mu (K)$ implies that $V$ satisfies $(3)$.

Wouldn't $\Lambda f\leq \mu (W)\Rightarrow\Lambda f\leq \mu (K)$ make sense only if $W\subset K$? But that is not the case since $K\subset W$... Also, how was the connection between $V$ and $K$ made, i.e. $K\subset V$? What is the connection between $W$ and $V$? And lastly why does $K\subset V$ and $\alpha<\mu (K)$ implies that $V$ satisfies $(3)$?

Any advice or hints would be appreciated. Thank you in advance.

Answer 1

The existence of $f$ follows from the definition of $\mu(V)$, as you said. Note, however, that this definition holds only for open sets and the definition for every set $E$ of $\mu$ is given by $$ \mu(E)=\inf\{\mu(V):E\subset V, V\quad \mathrm{open}\} $$ Using this, in particular we have $$ \mu(K)=\inf\{\mu(V):K\subset W, W\quad \mathrm{open}\} $$ The fact that $\Lambda f\leq \mu(W)$ for every $W$ such that $K\subset W$ implies that $\Lambda f$ is a lower bound of the set $\{\mu(V):K\subset W, W\quad \mathrm{open}\}$ so, by definition of infimum, we must have $\Lambda f\leq \mu(K)$. $W$ and $V$ doesn't need to have any connection, $W$ was introduced to show this last inequality.

Finally, note that Rudin defines $f\prec V$ if the support of $f$ lies in $V$. Since $K$ is the support of $f$, we must have $K\subset V$