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Borel set

Siron

Active member
Jan 28, 2012
150
How does one show the following ($\mathcal{R}$ denotes the one-dimensional Borel-set):
$$\mathcal{R} = \sigma(\mathcal{A})$$
with $\mathcal{A}=\{]a,b]|a,b \in \mathbb{Q}, a < b\}\cup {\emptyset}$

I know that $\mathcal{R}= \sigma(\{]a,b]|-\infty<a\leq b<\infty\})$ so I think $\mathcal{A} \subseteq \mathcal{R}$ and if I can show that $\mathcal{A}$ is a $\pi-$ systeme it follows by the $\pi-\lambda$ theorem that $\sigma(\mathcal{A})\subset \mathcal{R}$.

I could do the reverse implication the same way but then I need to prove that $\{]a,b]|-\infty<a\leq b<\infty\} \subset \sigma(\mathcal{A})$, but I'm struggling with this implication.

Anyone?
Thanks in advance.
 

Amer

Active member
Mar 1, 2012
275
what is
[tex] \sigma (A ) [/tex]