Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For those who have the Rudin: Real and complex analysis book at hand. I think I have an idea on step III of Riesz Rep. Theorem on page 43, but wasn't quite confident about my logic.

In the proof of Step III, we are trying to show that every open set $V$ satisfies $\mu(E) = \sup\{\mu(K) \mid K \subset E \ \quad K \ \mathrm{is} \ \mathrm{compact}\}$. From there, we can conclude that for all open sets $V$ in $X$, they lie in $\mathfrak{M}_F$ with $\mu(V) < \infty$.

My question is that since we are not assuming that $\mu(V)$ is finite, then I don't see where we conclude that $\mu(V) < \infty$. Near the end of the proof, I understood how Rudin concludes that $\alpha < \mu(K)$. Since we started with $\alpha < \mu(V)$, does this mean that $\mu(K)$ is close to $\mu(V)$? From Step II, $\mu(K) < \infty$ because it is in $\mathfrak{M}_F$. But I wasn't sure how to conclude that $\mu(V) < \infty$. Here's what I think how $\mu(V) < \infty$. Suppose that $\mu(V) = \infty$. Then any real number $\alpha < \mu(V)$ will satisfy this. Then from what we have shown in the proof, $\alpha < \mu(K)_{\alpha}$ for any $\alpha \in \mathbb{R}$, where $\mu(K)_{\alpha}$ depends on $\alpha$. This means that $\mu(K)_{\alpha} = \infty$ for arbitrarily large $\alpha$. This contradicts $\mu(K) < \infty$ in step II. This proof was done carefully and was made brief. But I wanted to understand the jump that Rudin made.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Rudin's "step III" has two assertions:

  • If $V$ is an open subset of $X$, then one has $$ \mu(V) = \sup \{\mu(K): K \subseteq V, K\text{ compact}\}. $$

  • Hence, if $V$ is an open subset of $X$ with $\mu(V) < \infty$, then $V$ is in $\mathfrak{M}_F$. (He words it: "hence, $\mathfrak{M}_F$ contains every open set $V$ with $\mu(V) < \infty$," but I have changed the ordering of the phrasing to make it clearer that $\mu(V) < \infty$ is a hypothesis and not a conclusion.)

So the second assertion is not that every open subset $V$ of $X$ satisfies $\mu(V) < \infty$. It is that if an open subset of $X$ does satisfy $\mu(V) < \infty$, then it is in $\mathfrak{M}_F$. Rudin does not explicitly deal with this in the proof at all, because once the first assertion is known, it is true by the way $\mathfrak{M}_F$ has been defined (it's the set of all subsets $E$ of $X$ satisfying both $\mu(E) < \infty$ and the condition mentioned in the first assertion).

So the work of step III is to fix an open set $V$ and show that the condition mentioned in the first assertion above holds.

This is the assertion that a particular number $x = \mu(V)$ is the supremum of a certain set $S = \{\mu(K): K \subseteq V, K\text{ compact}\}$ of real numbers.

Generally, to prove that a number $x$ is the supremum of a set $S$, you prove two things:

  1. The number $x$ is an upper bound of the set $S$.

  2. If $\alpha$ is a real number and $\alpha < x$, then $\alpha$ is not an upper bound of $S$.

As for the first bit: for any subset $K$ of $V$ at all, the inequality $\mu(K) \leq \mu(V)$ is an easy consequence of how $\mu$ is defined (he notes this earlier; when he refers to the "monotonicity" of $\mu$). In particular, for any compact subset $K$ of $V$, the inequality $\mu(K) \leq \mu(V)$ must hold. And this means that $\mu(V)$ is an upper bound for the set ${\mu(K): \text{$K \subseteq V$, $K$ compact}}$ .

Since there was so little to show here, Rudin doesn't make any explicit mention of it.

What Rudin labels as the entire proof of step III is just a proof of this second bit: that no smaller number than $\mu(V)$ is an upper bound for $\{\mu(K): K \subseteq V, K\text{ compact}\}$. Hence the $\alpha$ business: what he is showing in that paragraph is that if $\alpha$ is less than $\mu(V)$, then $\alpha$ cannot be an upper bound for the set $\{\mu(K): K \subseteq V, K\text{ compact}\}$ .

You should note that if $\mu(V) = \infty$, so that every real number $\alpha$ satisfies $\alpha < \mu(V)$, the argument he gives still works. It shows that no real number is an upper bound for $\{\mu(K): K \subseteq V, K\text{ compact}\}$. This means that $\sup \{\mu(K): K \subseteq V, K\text{ compact}\} = \infty$, and hence that $\mu(V) = \sup \{\mu(K): K \subseteq V, K\text{ compact}\}$ holds in this case also.

This is what many people love to hate about Rudin's textbook writing style: he makes a series of assertions, then he writes "Proof:", and then he gives an argument dealing with the hardest part of the proof of one of the assertions he just made. Even if you understand what he is doing, it takes a bit of getting used to.

I hope this helped.

share|improve this answer
1  
leslie, I fixed some display problems in your post (the preview and what is displayed when the answer is posted differ sometimes). There was the problem that most braces weren't displayed at all. Try to avoid nesting math environments inside \text: Write K\text{ compact} instead of \text{$K$ compact}. –  t.b. Apr 9 '12 at 7:04
    
I misunderstood the second assertion Rudin makes, but you've made it really clear. I didn't realize that was the hypothesis. It helped me a lot. Thank you! –  MathNewbie Apr 9 '12 at 7:11
    
@t.b. Thanks a lot! I never understood why that was happening. Now I do. (Oddly enough: for me at least, it was displaying correctly in the rendered "preview" of the solution to be posted. The fact that the posted solution wasn't matching the preview was completely baffling to me.) –  leslie townes Apr 14 '12 at 0:42
    
Apparently two different teams of developers work on the preview code and the code that's displayed on the posted version and that's why there are differences. By the way: I missed one instance that is still erroneously displayed at the end of the paragraph after the numbered list, but I didn't want to edit a second time only for this reason. –  t.b. Apr 14 '12 at 0:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.