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Prove that the following sets generate the Borel-$\sigma$-algebra $B(\mathbb{R})$

$$X_1 = \{ (-\infty,a), \, a \in \mathbb{R} \} \\X_2 = \{ (-\infty,b], \, b \in \mathbb{R} \}$$

We had defined $B(\mathbb{R})$ to be $\sigma$-algebra generated by all open subsets $O(\mathbb{R})$ of $\mathbb{R}$.

Here is my attempt:

In each case one inclusion is easy since $X_1, X_2 \subset O(\mathbb{R})$ which implies that $\sigma(X_1), \sigma(X_2) \subset \sigma(O(\mathbb{R})) = B(\mathbb{R})$.

For the other inclusion, we need only to show that $O(\mathbb{R}) \subset \sigma(X_i)$ since this would imply that $B(\mathbb{R})\subset \sigma( \sigma(X_i) ) = \sigma(X_i)$.

First let $A\in O(\mathbb{R})$ be an arbitrary open set. We may write it as $$A=\bigcup_{i\in \mathbb{N}} (x_i, y_i)$$ where each $x_k, y_k \in \mathbb{R}$. All these intervals are in $X_1$ and since $\sigma(X_1)$ is closed under countable unions, $A$ is an element of $\sigma(X_1)$.

Is this correct? How do I show the inclusion $B(\mathbb{R}) \subset \sigma(X_2)$?

Thank you!

J126
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rehband
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2 Answers2

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Just as you proved that $B(\mathbb R)\subset\sigma(X_1)$ you can prove that $B(\mathbb R)\subset\sigma(Y_2)$ where $Y_2$ denotes the set of complements of sets in $X_2$, i.e. $Y_2=\{(b,+\infty):b\in\mathbb R\}\subset O(\mathbb R)$.

Since $\sigma$-algebras are closed under complements we have $\sigma(Y_2)=\sigma(X_2)$.

edit:

After a better look at your attempt: you are saying "All these intervals are in $X_1$..." That is not true since $X_1$ only contains intervals of the form $(-\infty,a)$. Then $\sigma(X_1)$ will contain intervals of the form $[c,a)=(-\infty,a)\cap(-\infty,c)^c$ and each open interval $(x,y)$ can be written as countable union of intervals $[x+\frac{1}{n},y)\in\sigma(X_1)$. This proves that $\sigma(X_1)$ contains each open interval $(x,y)$. This opens the door to proving that each open set is contained by $\sigma(X_1)$.

drhab
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  • Thank you! Btw., can we still write $A=\bigcup_{i\in \mathbb{N}} (x_i, y_i)$ for rational numbers $x_i, y_i$ since $\mathbb{Q}$ is dense in $\mathbb{R}$? – rehband Oct 27 '14 at 11:14
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    Yes. You can do it with $x_i,y_i\in\mathbb Q$. – drhab Oct 27 '14 at 11:26
  • Can you tell me what the reasoning for this is? I think I got it but I don't know how to formulate it properly. Is it because if $x_i$ is irrational we can chose a rational sequence $x_i^{(n)}$ that converges to $x_i$ as $n\to\infty$? – rehband Oct 27 '14 at 11:32
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    The intervals $(r,s)$ with $r,s\in\mathbb Q$ form a (countable) base for the open set. Any open interval $(x,y)$ can be written as (countable) union of the intervals $(r,s)\subseteq (x,y)$ where $r,s\in \mathbb Q$. Consequently any open set - wich is a countable union of open intervals - can be written as union of these intervals $(r,s)$ – drhab Oct 27 '14 at 11:49
  • The last equality in your comment is not correct. – drhab Oct 27 '14 at 11:52
  • Yeap, I noticed and deleted it :P Thanks for the explainations – rehband Oct 27 '14 at 11:53
  • Shouldn't the union be of $[x-1/n,y)$? – Brofessor Aug 05 '17 at 04:37
  • @Brofessor No. $\bigcup_{n=1}^{\infty}[x-1/n,y)=[x-1,y)\neq[x,y)$ – drhab Aug 05 '17 at 07:15
  • I am referring to "each open interval $(x,y)$ can be written as countable union" $\bigcup_{n=1}^{\infty}[x-1/n,y)=(x,y)$ (as it is a union not so much an intersection) – Brofessor Aug 05 '17 at 08:57
  • @Brofessor This equality is false. Note that $[x-1,y)$ is a subset of the LHS and $x-1\notin(x,y)$. – drhab Aug 05 '17 at 11:30
  • I am mistaken! thank you @drhab – Brofessor Aug 05 '17 at 18:54
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As long as you can assume that those open intervals are in $X_1$, your proof is good. If not, you should prove that. To show that $B(\mathbb{R})\subseteq \sigma(X_2)$, you can show that $\sigma(X_1)\subseteq \sigma(X_2)$. Let $\{x_n\}$ be a sequence converging to $a$, such that $x_n< a$ for each $n$. Then $$ (-\infty,a)=\bigcup_{n=1}^\infty (-\infty,x_n]. $$

J126
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