Show that $\mathcal{J}=\{[a, a+1) \mid a \in \mathbb{R}\}$ generates the Borel set of $\mathbb{R}$ Problem Show that $\sigma(\mathcal{J})=\mathcal{B}(\mathbb{R})$, $\mathcal{J}=\{[a, a+1) \mid a \in \mathbb{R}\}$.

I'm using the following lemma:
Let $G,G´\subset P(X)$. $\sigma (G) = \sigma (G´) \iff G \subset \sigma (G´) \: \wedge \: G´\subset \sigma (G)$.

My solution
$\mathcal{J} \subset \mathcal{B}(\mathbb{R})$: It is known that the sets of half-open rectangles $\mathscr{J}=\{ \left[ b,c \right):b<c \in \mathbb{R} \}$ in $\mathbb{R}$ generates the Borel set on $\mathbb{R}$. So $\mathcal{J} \subset \mathscr{J} \subset \mathcal{B}(\mathbb{R}) \implies \mathcal{J} \subset \mathscr{B}(\mathbb{R})$ $\square$.
$\mathscr{J}\subset\sigma(\mathcal{J})$: My idea is that I will like to construct this type of interval $\left[ b,c \right),b<c \in \mathbb{R}$ using $ \left[ a,a+1 \right),a \in \mathbb{R}$ using all the set operations available for which a $\sigma$-algebra is closed under. From $ \left[ a,a+1 \right),a \in \mathbb{R}$ I find it hard to construct an interval with an arbitrary right interval point instead of $a+1$.
Hints and help are welcome.
 A: Here is one way to do it:
I leave some details to the OP

*

*First notice that the sigma algebra $\mathcal{A}$ generated by the intervals $[a,a+1)$, $a\in\mathbb{R}$ is contained in the Borel $\sigma$-algebra (why?)


*Notice that dir any $b\in\mathbb{R}$, $(-\infty,b)=\bigcup^\infty_{n=1}[b-n,b-n+1)$ is in $\mathcal{A}$.


*Then, for any $a\in\mathbb{R}$,  $(-\infty,a]=\bigcap^\infty_{k=1}(-\infty,a+\tfrac{1}{k})$ is also in $\mathcal{A}$. Consequently, $\mathbb{R}\setminus(-\infty,a]=(a,\infty)$ is in $\mathcal{A}$. This means that $(a, b)=(a,\infty)\cap(-\infty,b)$, for all $a<b$, is in $\mathcal{A}$.


*Thus $\mathcal{A}$ contains all open intervals (finite and infinite) and so all open sets in $\mathbb{R}$ (every open set in $\mathbb{R}$ is the countable union of disjoint open intervals.


*So $\mathcal{A}$ and the Borel sigma algebra are the same.
A: For $a<b$ with $b-a\leq1$, we have $[a,b)=[a,a+1)\setminus[b,b+1)\in\sigma(\mathcal{J})$.
Note that every interval of the form $[a,b)$ can be written as finite
union of intervals of the form $[c,d)$ with $d-c\leq1$. Therefore
$[a,b)\in\sigma(\mathcal{J})$. Next, $(a,b)=\cap_{n}[a+\frac{1}{n},b)\in\sigma(\mathcal{J})$.
Since every open subset of $\mathbb{R}$ is a countable union of open
intervals, $\sigma(\mathcal{J})$ contains every open subset of $\mathbb{R}$.
It follows that $\sigma(\mathcal{J})\supseteq\mathcal{B}(\mathbb{R})$.
The reverse inclusion is trivial.
A: Let $G=\mathcal J$ and let $G'$ be the set of bounded open intervals.  I assume you know that $\sigma(G')=\mathcal B(\Bbb R).$
1.If $g'=(x,y)\in G'$ with $y>x+1$ then $g'=\cup \{[z,z+1): z\in \Bbb Q\cap (x,y-1]\}\in \sigma (G).$


*If $g'=(x,y)\in G'$ with $y\le x+1$ then $(y-2,y)$ and $(x,x+2)$ both belong to $\sigma (G)$ by 1. So $g'=(y-2,y)\cap (x,x+2)\in \sigma (G).$
Therefore $G'\subset \sigma (G).$
