# Rudin's Proof on Riesz Representation Theorem

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 $$(3)$$?

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 $K\subset V$