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I think I should show that $\beta$ and $A$ contain both open intervals and intervals of the form $[a,b)$, but I have no idea how to proceed with either of those.

Xam
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Vinny Chase
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2 Answers2

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Half-open intervals $\mathcal{A}=\{[a,b):a<b\}$ are Borel sets. Thus, $\sigma(\mathcal{A})\subset \mathcal{B}_{\mathbb{R}}$. On the other hand, any open set in $\mathbb{R}$ can be approximated by a countable union of the intervals from $\mathcal{A}$ which implies that $\mathcal{B}_{\mathbb{R}}\subset\sigma(\mathcal{A})$.

  • How does the fact that any open set can be written as a countable disjoint union of open intervals imply that $\beta \subset \sigma(A)$? I'm new to $\sigma-algebras$ so I'm pretty lost – Vinny Chase Sep 21 '17 at 20:44
  • It means that open sets are in the $\sigma$-algebra generated by $\mathcal{A}$. Thus, the $\sigma$-algebra generated by all open sets ($\beta$) is contained in the first one. https://proofwiki.org/wiki/Definition:Sigma-Algebra_Generated_by_Collection_of_Subsets –  Sep 21 '17 at 21:03
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Let $\mathcal{I}$ be the collection of all open intervals, and $\mathcal{A}$ be the collection of all half open-intervals $[a,b)$. We shall prove $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$ and $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.

Open intervals are open sets so $\mathcal{I} \subseteq \mathcal{B}_\mathbb{R} \implies \sigma(\mathcal{I}) \subseteq \mathcal{B}_\mathbb{R}$.

Take an open set $U \subseteq \mathbb{R}$. For $x \in U$ there exists an open interval $I_x \subset U$ such that $x \in U$. Take $I_x$ to be the largest such interval. We have $U = \bigcup_{x\in U} I_x$. The intervals $I_x$ are disjoint, otherwise they wouldn't be the largest. Since a union of open intervals can be disjoint only if it is at most countable (since each of them contains a distinct rational number), there exists countably many intervals $I_{x_{n}}$ such that $U = \bigcup_{n=1}^\infty I_{x_{n}}$. Thus, $U \in \sigma(\mathcal{I})$.

Hence, $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$.

Now, take an open interval $(a, b) \in \mathcal{I}$ and notice $$ (a, b) = \bigcup_{n=1}^\infty \underbrace{\left[a +\frac{1}{n}, b\right)}_{\in \mathcal{A}} \in \sigma(\mathcal{A})$$

so $\mathcal{I} \subseteq \sigma(\mathcal{A})\implies \sigma(\mathcal{I}) \subseteq \sigma(\mathcal{A})$.

Similarly, take $[a,b) \in \mathcal{A}$ and notice:

$$ [a, b) = \bigcap_{n=1}^\infty \underbrace{\left(a -\frac{1}{n}, b\right)}_{\in \mathcal{I}} \in \sigma(\mathcal{I})$$

so $\mathcal{A} \subseteq \sigma(\mathcal{I})\implies \sigma(\mathcal{A}) \subseteq \sigma(\mathcal{I})$.

Hence, $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.

Finally, we can conclude $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{A})$.

mechanodroid
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  • Thanks! Could you explain why the Ix are disjoint in a little more detail? Why would them not be contradict them being largest? – Vinny Chase Sep 21 '17 at 20:34
  • Also, why does $\mathcal(U)$ $\subset$ $\sigma(I)$ imply $\beta$ = $\sigma(I)$ – Vinny Chase Sep 21 '17 at 20:39
  • @VinnyChase Assume $I_x \cap I_y \ne \emptyset$ for some $x \ne y$. But, this means that $I_x \cup I_y$ is also an open interval contained in $U$ which contains $x$. Since $I_x \subsetneq I_x \cup I_y$, this is a contradiction with the fact that $I_x$ is the largest interval in $U$ which contains $x$. Have a look here. – mechanodroid Sep 22 '17 at 08:36
  • I assumed you defined $\mathcal{B}\mathbb{R}$ as $\sigma(\mathcal{U})$, where $\mathcal{U}$ is the collection of open subsets of $\mathbb{R}$. Then I showed $\mathcal{U} \subseteq \sigma(I)$ so $\mathcal{B}\mathbb{R} = \sigma(\mathcal{U}) \subseteq \sigma(I)$. – mechanodroid Sep 22 '17 at 08:40