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The proof is from Rudin's Real and Complex Analysis.

The Theorem states: Every open set $V$ 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 $E$ 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 $V$.

What I understand from the proof/paraphrasing:

Suppose that we have a measure $\mu(V)$, where $V$ is open. $\mu(V)$ must be greater than some real number $\alpha$. Furthermore, there should exist a function $f$ whose support lies in $V$ such that the $\Lambda f>\alpha$. Now, for any open set $W$ that contains the support $K$ of $f$, $$\Lambda f\leq \mu(W)$$ Note that for $K$ is such that $K\subset K$ and contains the support of $f$. Thus, $$\Lambda f\leq \mu(K)$$ as well. I don't understand anything else that comes after.

Help: Why is $\alpha$ needed? What does it do? Why is the inequality $$\alpha<\mu(K)$$ needed? How does it imply that $$\mu(E)=\sup\,\{\mu(K):K\subset E,\,K\,\,\text{compact}\}$$ Also, what is the proof referring to by the word "this" in "This exhibits a compact..."?

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

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    $\begingroup$ You used the letter $V$ twice where it seems to be $E$. $\endgroup$ – Thibaut Dumont May 9 '15 at 11:03
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It's a squeeze technique. We know $\mu(K)\le\mu(V)$ since $K \subset V$, and you now have $\alpha <\mu(K)\le\mu(V)$ with $\alpha$ arbitrary, so let it tend to $\mu(V)$.

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