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I'm preparing for my exam. Can anyone help me in this matter, is confusing to me

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thank you very much.

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closed as off-topic by user223391, levap, anomaly, Nikunj, Davide Giraudo May 31 '16 at 21:11

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  • $\begingroup$ I'm assuming $\mathcal{P}(\mathbb{Q})$ refers to the Borel $\sigma$-algebra on $\mathbb{Q}$? $\endgroup$ – Jason Jan 30 '15 at 5:01
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    $\begingroup$ Do you know what $\mathcal{P}(\mathbb{Q})$ is? Do you recognise that $\mathbb{Q} \cap (a,b) = \{(a,b) : a < b, a,b \in \mathbb{Q}\}$ and that for any $r \in \mathbb{Q}$, we have $\{r\} = \cap_{n=1}^\infty (r-n^{-1},r]$? Can you prove this cannot be accomplished with finite unions and complementation? $\endgroup$ – snar Jan 30 '15 at 5:04
  • $\begingroup$ What are you stuck on? All of these can be verified directly from the relevant defintions. $\endgroup$ – anomaly May 31 '16 at 18:05
  • $\begingroup$ @Jason $\mathcal P(\Bbb Q)$ is the set of all the subsets of $\Bbb Q$. Which is incidentally the Borel $\sigma$-algebra of $\Bbb Q$ as well, but that's kind-of point (b) of the question. $\endgroup$ – user228113 May 31 '16 at 19:12
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Part b:

Here $\mathcal{P}(\mathbb Q)$ is simply the powerset of $\mathbb Q$. Thus we are trying to show that every subset of $\mathbb Q$ can be obtained by countably many union and complement operations on the sets $(a,b]$.

First observe that any singleton $\{r\}$ can be obtained in the manner, since $$ \{r\}=\bigcap_{\substack{s<r\\s\in \mathbb Q}}(s,r]. $$ Next observe that for any $R\subset \mathbb Q$, we have $$ R=\bigcup_{r\in R}\{r\}. $$ Consequently $\mathcal{P}(\mathbb Q)$ is the $\sigma$-algebra generated by the sets $(a,b]$.

Part c:

Define $\mu(\{r\})$ to be any number you like, for each $r\in\mathbb{Q}$ and extend to all of $\mathcal{P}(\mathbb{Q})$ by $\sigma$-additivity. You get $\mu((a,b])=\infty=\mu_0((a,b])$.

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Let $\mathcal{A}$ be the collection of finite unions of sets of the form $(a,b]\cap\mathbb{Q}$ where $-\infty \leq a < b\leq \infty$

a.) $\mathcal{A}$ is an $\sigma$-algebra on $\mathbb{Q}$

b.) The $\sigma$-algebra generated by $\mathbb{A}$ is $\mathcal{P}(\mathbb{Q})$

c.) Define $\mu_0$ on $\mathcal{A}$ by $\mu_0(\emptyset) = 0$ and $\mu_0(A) = \infty$. Then $\mu_0$ is a premeasure on $\mathcal{A}$, and there is more than one measure on $\mathcal{P}(\mathbb{Q})$ whose restriction to $\mathcal{A}$ is $\mu_0$.

Proof a.) - Let $\varepsilon = \{(a,b]\cap \mathbb{Q}: a,b\in\overline{\mathbb{R}}\}$. Clearly $\emptyset = (0,0]\cap\mathbb{Q}\in\varepsilon$. If $(a_1,b_1]\cap\mathbb{Q}\in\varepsilon$ and $(a_2,b_2]\cap\mathbb{Q}\in\varepsilon$ then their intersection is $(\max\{a_1,a_2\},\min\{b_1,b_2\})\cap\mathbb{Q}\in \varepsilon$. Moreover, the complement of $(a,b]\cap\mathbb{Q}$ is $(-\infty,a]\cap\mathbb{Q}\cup (b,\infty]\cap\mathbb{Q}$ which is a disjoint union of elements of $\varepsilon$ if $a\leq b$. If $a>b$ then the complement of $(a,b]\cap\mathbb{Q}$ is $(-\infty, \infty]\cap\mathbb{Q}\in\varepsilon$. This shows that $\varepsilon$ is an elementary family of subsets of $\mathbb{Q}$, so the collection of finite disjoint unions of members of $\varepsilon$ is an algebra by proposition 1.7. Let $(a_1,b_1]\cap\mathbb{Q}\in\varepsilon$ and $(a_2,b_2]\cap\mathbb{Q}\in\varepsilon$ be not disjoint. Then, $(a_1,b_1]\cap\mathbb{Q}\cup (a_2,b_2]\cap\mathbb{Q} = ((a_1,b_1]\cup (a_2,b_2])\cap\mathbb{Q}$. So we need to figure out what is $(a_1,b_1]\cup (a_2,b_2]$ which we have 4 different outcomes: \begin{align*} (a_1,b_1]\cup (a_2,b_2] =\\ &1. (a_1,b_2] \ \ \text{if} \ \ a_1\leq a_2\leq b_1 < b_2\\ &2. (a_1,b_1] \ \ \text{if} \ \ a_1\leq a_2< b_2 \leq b_1\\ &3. (a_2,b_2] \ \ \text{if} \ \ a_2\leq a_1\leq b_1 \leq b_2\\ &4. (a_2,b_1] \ \ \text{if} \ \ a_2\leq a_1< b_2 \leq b_1\\ \end{align*} which are all in $\varepsilon$ thus $\mathcal{A}$ is the collection of finite disjoint unions of members of $\varepsilon$, so $\mathcal{A}$ is a $\sigma$-algebra.

Proof b.) - Suppose that $r\in\mathbb{Q}$. Then $$\{r\} = \bigcap_{n\geq 1}((r-1/n,r]\cap\mathbb{Q})\in M(\mathcal{A})$$ If $X\subset \mathbb{Q}$, then $X$ is countable, say $X = \{x_1,\ldots\}$. But $\{x_j\}\in M(\mathcal{A})$ for all $j\geq 1$. So $$X = \bigcup_{j\geq 1}\{x_j\}\in M(\mathcal{A})$$ Therefore $M(\mathcal{A}) = \mathcal{P}(\mathbb{Q})$

Proof c.) - It is clear that $\mu_0$ is a premeasure on $\mathcal{A}$. The induced outer measure is $$\mu^*(E) = \inf\{\sum_{1}^{\infty}\mu_0(E_j):E_j\in\mathcal{A},E\subset \bigcup_{1}^{\infty}E_j\}$$ it follows that from theorem 1.14 that $\mu = \mu^*$ is a measure on $\mathcal{P}(\mathbb{Q})$ which extends $\mu_0$.

If $\nu$ is a counting measure on $\mathcal{P}(\mathbb{Q})$ then $\nu(\emptyset) = 0 = \mu_0(\emptyset)$. And for $a<b$, $E = (a,b]\cap\mathbb{Q}$ has infinite cardinality, so $\nu(E) = \infty = \mu_0(E)$ and therefore $\nu|\varepsilon = \mu_0|\varepsilon$. By finite additivity of $\nu$, it follows that $\nu(E) = \infty = \mu_0(E)$ for all $\emptyset \neq E\in\mathcal{A}$. So that $\nu|\mathcal{A} = \mu_0$. But $\nu\{1\} = 1 < \infty = \mu(\{1\})$, so $\nu$ is a different measure on $\mathcal{P}(\mathbb{Q})$ which extends $\mu_0$.

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