I want to prove that $ \sigma \{ \big( \frac{1}{n+1} ,\frac{1}{n} \big] : n \in \mathbb{N} \} = \mathbb{B}(0,1] $
i.e, the sigma algebra generated by $\{ \big( \frac{1}{n+1} ,\frac{1}{n} \big] : n \in \mathbb{N} \}$ is equal to the Borel sigma algebra of $(0,1]$
$\mathbb{B} (0,1] $ is defined by $\mathbb{B} (0,1] = \{ (0,1] \cap B : B \in \mathbb{B}( \mathbb{R} ) \} $, where $\mathbb{B}( \mathbb{R} )$ is the Borel sigma algebra of $ \mathbb{R} $
Here's my attempt:
$\subset$:
Let $ n\geq 1 $ be an integer, then $ 1/n \leq 1 $. We also have $0 < 1/(n+1)$. This implies that
\begin{equation} \Big(\frac{1}{n+1},\frac{1}{n} \Big] \subset (0,1] \end{equation}
we know that sets of the form $(a,b] \in \mathbb{B}( \mathbb{R} ) $ for any $a \leq b$. Then
\begin{equation} \Big(\frac{1}{n+1},\frac{1}{n} \Big] = (0,1] \cap \Big(\frac{1}{n+1},\frac{1}{n} \Big] \in \mathbb{B}(0,1] \end{equation} This implies that
\begin{equation} \Big\{ \Big( \frac{1}{n+1} ,\frac{1}{n} \Big] : n \in \mathbb{N} \Big\} \subset \mathbb{B}(0,1] \end{equation} since $ \mathbb{B}(0,1] $ is a sigma algebra, we get \begin{equation} \sigma\Big\{ \Big( \frac{1}{n+1} ,\frac{1}{n} \Big] : n \in \mathbb{N} \Big\} \subset \mathbb{B}(0,1] \end{equation}
$\supset$:
I want to show that $ \mathbb{B}(0,1] \subset \sigma \{ \big( \frac{1}{n+1} ,\frac{1}{n} \big] : n \in \mathbb{N} \} $
Let $A \in \mathbb{B}(0,1] $ then $A = (0,1]\cap B $ where $B \in \mathbb{B}( \mathbb{R} ) $. This implies that $A \subset (0,1]$ and $A \subset B$. But at this point I'm stuck, I don't know how to proceed.
My questions are:
- Do you know if these sets are equal? if so,
- Do you have any hint that could help me to prove the equality?
- Do you know some references where something similar to this problem is done?
Thanks in advanced.