I have the following (simple) homework:
Let $\mathcal{F}$ be $\{[a,a+1)|a \in \mathbb{R}\}$. Show that $\sigma(\mathcal{F})=\mathcal{B(\mathbb{R})}$
My idea was show that a borel set [a,b) is:
$[a,b)=$($\cup_{n\in\mathbb{N}} [a+n,a+n+1)$ ) $ \bigcap$ ($(\cup_{n\in\mathbb{N}} [b+n,b+n+1))^c$ )
But can I choose to show $[a,\infty)$ instead or is it the correct approach, input would be appreciated
Your argument shows that each $[a,b)$ is contained in $\sigma(\mathcal{F})$. Now, you can prove that each $(a,b)$ is contained in $\sigma(\mathcal{F})$, since $(a,b) = \cup_{n} [a + \frac{1}{n} , b)$.
To solve your problem directly, you need to prove that each open set is in $\sigma(\mathcal{F})$. Then $\sigma(\mathcal{F}) \supseteq \mathcal{B}(\mathbb{R})$, since the Borel sigma-algebra is generated by every open sets in $\mathbb{R}$.
Hint: each open set is a countable union of open intervals.