Trying to understand this very long theorem of which I think good understanding is very educational. I am going through all the steps specifically now there's a subtlety in the conclusion of step II which I cannot get.

I am sure is a silly thing. I'll write the proof for reference

Step II : If $K$ is compact, then $K \in \mathcal{M}_F$ and $$ \mu(K) = \inf\left\{\Lambda f : K \prec f \right\} \;\;\;\; (7) $$ This implies assertion (b) of the theorem.

Proof: If $K \prec f$ and $0 < \alpha < 1$, let $V_\alpha = \left\{x : f(x) > \alpha \right\}$. Then $K \subset V_\alpha$ and $\alpha g \leq f$ whenever $g \prec V_\alpha$. Hence $$ \mu(K) \leq \mu(V_\alpha) = \sup \left\{ \Lambda g : g \prec V_\alpha \right\} \leq \alpha^{-1} \Lambda f $$ Let $\alpha \to 1$ to conclude that $$ \mu(K) \leq \Lambda f \;\;\;\; (8) $$ Thus $\mu(K) < \infty$. Since $K$ evidently satisfies (3), $K \in \mathcal{M}_F$. If $\epsilon > 0$, there's $V, K \subset V$ with $\mu(V) < \mu(K) + \epsilon$. By Urysohn's lemma $K \prec f \prec V$ for some $f$. Thus $$ \Lambda f \leq \mu(V) < \mu(K) + \epsilon $$ which, combined with (8), gives (7)

Here's the question or maybe clarification... combining (7) with (8) provides to me that

$$ \mu(K) \leq \Lambda f \leq \mu(K) + \epsilon \;\;\;\; (9) $$

and bedause $\epsilon > 0$ is arbitrary we end up with

$$ \mu(K) = \Lambda f $$

However in (8) the relationship $\mu(K) \leq \Lambda f$ holds for any $f$ with $K \prec f$ and specifically we have

$$ \mu(K) \leq \inf \left\{ \Lambda f : K \prec f \right\} $$

On the other hand for $9$ the $f$ used depends on the open $V$ chosen therefore to me the combination of (8) and (9) to get (7) should be done as

$$ \mu(K) \leq \inf \left\{\Lambda f : K \prec f \right\} \leq \Lambda f \leq \mu(K) + \epsilon \Rightarrow \mu(K) \leq \inf \left\{\Lambda f : K \prec f \right\} \leq \mu(K) + \epsilon $$

and as $\epsilon \to 0$ we have

$$ \mu(K) = \inf \left\{\Lambda f : K \prec f \right\} $$

Is this the right conclusion?

  • $\begingroup$ There is an error in the quoted reference. Rudin says $\alpha g \leq f$ and $\mu(K) \leq \mu(V_\alpha)$, whereas you have the reverse inequality. $\endgroup$
    – shoteyes
    Jul 19, 2021 at 19:41
  • $\begingroup$ @shoteyes Should be ok now? Thank you $\endgroup$ Jul 20, 2021 at 10:47
  • $\begingroup$ what does $\prec$ mean? Sorry but I am not aware of that sign. Thanks :) $\endgroup$
    – 0m3g4
    Jul 20, 2021 at 10:50
  • 1
    $\begingroup$ @Algebrology if $K$ is compact then $K \prec f$ means $f$ has compact support, $0 \leq f \leq 1$ and $f(x) = 1$ for every $x \in K$. If $V$ is open the $f \prec V$ means the support of $f$ is in $V$ and $0 \leq f \leq 1$. $\endgroup$ Jul 20, 2021 at 10:54
  • $\begingroup$ @user8469759 Thanks for your help :) $\endgroup$
    – 0m3g4
    Jul 20, 2021 at 11:02

1 Answer 1


Your conclusion is correct but perhaps overly complicated.

The first part of the proof says that for every $f$ such that $K \prec f$, we have $\mu(K) \leq \Lambda f$. In other words, $\mu(K)$ is a lower bound for the set $\{\Lambda f\colon K \prec f\}$.

The second part of the proof says that for every $\varepsilon > 0$, there exists an $f$ such that $K \prec f$ and $\Lambda f < \mu(K) + \varepsilon$. In other words, $\mu(K) + \varepsilon$ is not a lower bound for the set $\{\Lambda f\colon K \prec f\}$.

So $\mu(K)$ is a lower bound, and everything that’s greater than $\mu(K)$ is not a lower bound. This is precisely what it means to be the greatest lower bound. Hence

$$\mu(K) = \inf{\{\Lambda f\colon K \prec f\}}.$$

  • $\begingroup$ Is it really over complicating? I think I am saying the same thing you're =D. $\endgroup$ Jul 20, 2021 at 10:39
  • $\begingroup$ @user8469759 Of course we’re saying the same thing; we’re both right! I just thought I could show you a way of saying it in words rather than chains of inequalities. I happen to prefer that kind of explanation, but the important thing in the end is that you understand why it’s true :) $\endgroup$
    – shoteyes
    Jul 20, 2021 at 11:29
  • $\begingroup$ Wasn't sure I missed something else, or I was actually overcomplicating "mathematically". $\endgroup$ Jul 20, 2021 at 11:40
  • $\begingroup$ Sometimes Rudin doesn't really explain what he's trying to prove and which technique he's using. There's a proof his functional analysis book where I didn't realize he was using a contrapositive argument, which drove me crazy for a week. Another thing I don't quite like about his books is the lack of examples, proof are amazing but he could've done more examples to show what sort of things you can do with his theorem. This version of Representation theorem is used to define the Lebesgue measure in the same chapter, but other than that I don't see many "toy" applications that would useful. $\endgroup$ Jul 20, 2021 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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