I've been annotating the steps in Riesz Rep. Theorem. So far, I have almost all of them. I just have six questions, some are short questions. So these questions are for anyone who has Rudin Real and Complex analysis on hand.

(1) My first question starts on page $44$ for step IV. We start with two disjoint compact sets $K_1, K_2$. I'm not really clear on what $f(x) = 0$ on $K_2$. My reasoning is that we can an open set $V$ such that it contains $K_1$ and $V \cap K_2 = \emptyset$. So using Urysohn's lemma, we get $K_1 \prec f \prec V$ for some $f \in C_c(X)$. The support of $f$ lies in $V$, and since $K_2 \cap V = \emptyset$, $f(x) = 0$ for all $x \in K_2$. Are we guaranteed that we can find such open set $V$?

(2) My next question is on the same step but at equation $(13)$. So far, it says $(9)$ shows that $(13)$ holds. I don't see how this is obvious. For me, i started with step I, used that $\mu(E) \leq \sum_{1}^{\infty}\mu(E_i)$. I just use the fact that there must be infinite number of zero sets or else $\mu(E) < \infty$ is violated. Also there must be finite number of non-zero sets, so is that why we get $(13)$?

(3) On page 46 Step $X$. Rudin says, "Clearly, it is enough to prove this for real $f$". I don't have a complex analysis background but is he insinuating that the case for $f$ being complex is almost the same?

(4) On the same page, between equation $(18)$ and $(19)$, Rudin mentions that the sets $E_i$ are therefore disjoint Borel sets whose union is $K$. I understand this paragraph except the part that $E_i$ are Borel sets. I'm thinking that I have to verify that $E_i$ are either closed or open. We have that $f$ is continuous so the pre-image of an open set is open. But I am having trouble getting to verify that fact.

(5) In the same paragraph, Rudin says "There are open sets $V_i \supset E_i$ such that $\mu(V_i) < \mu(E_i) + \frac{\epsilon}{n}$" equation $(19)$ and such that $f(x) < y_i + \epsilon$ for all $x \in V_i$. Using equation $(2)$, I understand that $\mu(E_i) + \epsilon > \mu(V)$ for some open set $V$ such that $V \supset E$. Why does Rudin define equation $(19)$ as that way (the epsilon term divided by n) and was wondering why $f(x) < y_i + \epsilon$ for $x \in V_i$ <--- for this part, I can get a $V_i$ satisfying this inequality but dont see how it also satisfy $(19)$

(6) Same page, at the bottom, Rudin says that Step II shows that $\mu(K) \leq \Lambda(\sum h_i) = \sum \Lambda h_i$. This isn't obvious at all. I looked at $\mu(K) = \inf\{\Lambda f \mid K \prec f\}$ but we have that $h_i \prec V_i$. I tried showing that $K \prec \sum h_i$. But what seems to be the trouble is verifying that $0 \leq \sum h_i \leq 1$ for all $x \in X$

Thanks a bunch!


A couple of A few remarks:

  1. $V= (K_2)^c$.

  2. Since $\mu(E) = \sum_{i=1}^\infty \mu(E_i) \lt \infty$ you have that $\lim_{N \to \infty} \sum_{i=N+1}^\infty \mu(E_i) = 0$. Take $N$ so large that this last sum is smaller than $\varepsilon \gt 0$.

  3. Since both sides of the equation $\Lambda f = \int f\,d\mu$ are complex linear in $f$, you can decompose $f$ as $f = \operatorname{Re}{f} + i \operatorname{Im}{f}$ and get from the equality for real functions that the one for complex functions holds as well: $$ \Lambda f = \Lambda \operatorname{Re}{f} + i \Lambda \operatorname{Im}{f} = \int \operatorname{Re}{f} \, d\mu + i \int \operatorname{Im}{f}\,d\mu = \int f\,d\mu, $$ as desired.

  4. Since $f$ is continuous, $O_i = \{x\,:\,y_{i-1} \lt f(x)\}$ is open and $F_i = \{x\,:\,f(x) \leq y_i\}$ is closed. Thus both are Borel sets and by definition $E_i = O_i \cap F_i$, so $E_i$ is a Borel set, too.

    More generally: if $f:X \to \mathbb{R}$ is Borel measurable and $B \subset \mathbb{R}$ is Borel then $f^{-1}B \subset X$ is Borel. To see this, note that $\Sigma = \{A \subset \mathbb{R}\,:\,f^{-1}A \text{ is Borel}\}$ is a $\sigma$-algebra containing the half-open intervals $(a,\infty)$ by definition of measurability, hence it contains all the open intervals, hence it contains the Borel sets. Clearly $(y_{i-1},y_i]$ is a Borel set, hence so is $E_i = f^{-1}(y_{i-1},y_i]$.

  5. There are $n$ sets $E_1,\ldots,E_n$ in this construction. The division by $n$ is simply to achieve that $\mu(V_1 \cup \cdots \cup V_n) \leq \sum_{i=1}^n \mu(V_i) \leq \sum_{i = 1}^n (\mu(E_i) + \varepsilon/n) = \varepsilon + \sum_{i=1}^n \mu(E_i) = \varepsilon + \mu(E)$ and this allows to eliminate $n$ in the very last estimate on page 47.

    To get a $V_i$ as the one required take the intersection of the $V_i$ you got with the open set $\{x\,:\,f(x) \lt y_i + \varepsilon\} \supset E_i$.

  6. The functions $h_i$ form a partition of unity on $K$ subordinate to $V_i$ and (the proof of) Theorem 2.13 (where the $h_i$ come from) gives you exactly what you need.

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  • 5
    $\begingroup$ That’s not a couple; that’s a group marriage! :-) $\endgroup$ – Brian M. Scott Apr 4 '12 at 6:34
  • 2
    $\begingroup$ @MathNewbie: A function is continuous if and only if pre-images of open sets are open (by definition) and since $f^{-1}[U^c] = [f^{-1}U]^c$ this is equivalent to requiring that $f^{-1}A$ is closed whenever $A$ is closed. For (2), suppose that $\mu(E_i \cap E_j) \gt 0$ for some $i \neq j$. Then $\mu(E_i \cap E_j) + \mu(E_i \cup E_j) = \mu(E_i) + \mu(E_j)$ shows that $\mu(E_i \cup E_j) = \mu(E_i) + \mu(E_j) - \mu(E_i \cap E_j)$ which implies that the sum on the right hand side is strictly larger than the one on the left hand side: $\endgroup$ – t.b. Apr 4 '12 at 10:32
  • 2
    $\begingroup$ $\mu(E) = \mu\left( (E_i \cup E_j) \cup \bigcup_{k \neq i,j} E_k\right) \leq \mu(E_i \cup E_j) + \sum_{k \neq i,j} \mu E_k \leq - \mu(E_i \cap E_j) + \sum_k \mu(E_k)$. Thus, if equality holds, the sets $N_{ij} = E_i \cap E_j$ for $i \neq j$ must be null-sets and taking $N = \bigcup_{i \neq j} N_{ij}$ (which is a null-set) and replacing $E_i$ by $E_i \smallsetminus N$ you get what you want. $\endgroup$ – t.b. Apr 4 '12 at 10:34
  • 2
    $\begingroup$ @MathNewbie: I probably misread what you wrote, already assuming that $\mu$ is a measure and arguing by contradiction that equality $\mu(E) = \sum_{1}^\infty \mu(E_i)$ implies that $\mu(E_i \cap E_j) = 0$. Can you ask question (2) again? What do you want to assume on $\mu$, etc? The equality $\mu(E_i \cap E_j) + \mu(E_i \cup E_j) = \mu(E_i) + \mu(E_j)$ follows from the disjoint decompositions $E_i = (E_i \smallsetminus (E_i \cap E_j)) \cup (E_i \cap E_j)$ and $E_i \cup E_j = (E_i \smallsetminus (E_i \cap E_j)) \cup E_j$ and using that a measure is additive on disjoint sets. $\endgroup$ – t.b. Apr 4 '12 at 11:13
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    $\begingroup$ @MathNewbie: Okay I think now I understand: Equation $(9)$ is already proved for pairwise disjoint sets $E_i \in \mathfrak{M}_F$ and $E = \bigcup E_i$. The point of $(13)$ is: we know that $\mu(E) = \sum_{i=1}^\infty \mu(E_i) \lt \infty$, so the series on the right hand side is absolutely convergent. Since all summands are positive this means that $0 \leq \sum_{1}^N \mu(E_i) \leq \mu(E)$ and $s_N = \sum_{1}^N \mu(E_i)$ converges monotonically to $\mu(E)$. In other words, for every $\varepsilon \gt 0$ there is $N$ such that $0 \leq \mu(E) - s_N \lt \varepsilon$. Now re-arrange this to get $\endgroup$ – t.b. Apr 4 '12 at 11:45

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