Show $\sigma (A) = \mathscr{B} (\mathbb{R})$ So Im trying to find out how to show that $\sigma (A)= \mathscr{B}( \mathbb{R})$ given $A=\{[a,a+1)|a \in \mathbb{R}\}$.
Sigma algebras is new for me, so I am pretty unsure on were to start.
Maybe something about the families of open rectangles?
Could anyone maybe give a hint or an outline of how to work with a problem like this?
 A: First note that you can produce any $[a, b)$ from $A$, for any $ a  < b$ in $\mathbb{R}$. You'll just have to break it up into a bunch of cases:

*

*If $b-1 < a$ then $[a,  a+1) \cap [b-1, b) = [a, b)$


*If $a + 1 <b$ then you do a bunch of unions $[a, a+1) \cup [a+1, a+2) \cup ...$ until you get to some $k$ such that $a + k < b < a+k+1$ and then you apply case $1$ again. So that $[a, a+1) \cup [a+1, a+2) \cup ... \cup \big([a + k, a+k+1) \cap [b-1, b)\big) = [a, b)$.
Thus we know that $\sigma(A)$ has all the half open intervals. It's a pretty common fact that the half open intervals generate the Borel Sigma Algebra (if you don't know this, it would be a good exercise to try and prove it yourself).
Edit: This would show that $\sigma(A) \supseteq \mathcal{B}(\mathbb{R})$ but of course the other containment is trivial (since the Borel Sigma algebra already contains elements of the form $[a, a+1)$ as the intersections of open intervals $\cap_{n \geq 1}(a-\frac{1}{n}, a+1)$, this means that $A \subseteq \mathcal{B}(\mathbb{R})$ and so $\sigma(A) \subseteq \mathcal{B}(\mathbb{R})$ as well).
Let me know if you'd like to expand on any of the above.
