# Thread: Borel algebra over the real line

1. ## Borel algebra over the real line

Show that the Borel algebra over $\displaystyle \mathbb {R}$, denoted $\displaystyle \mathbb {B} _ \mathbb {R}$ is generated by the set of $\displaystyle \xi= \{ [a, \infty ) : a \in \mathbb {R} \}$

Set up:

I know that $\displaystyle \mathbb {B} _ \mathbb {R} = \mathbb {M} (U)$, where $\displaystyle \mathbb {M} (U)$ is the smallest $\displaystyle \sigma$-Algebra generated by U, and U is the family of all open sets in $\displaystyle \mathbb {R}$. So by a theorem that I know, if I can prove that $\displaystyle E \in \mathbb {M} (F)$, then I have $\displaystyle \mathbb {M} (E) \subset \mathbb {M} (F)$. My plan is to prove that $\displaystyle \xi \in \mathbb {M}(U)$ and vice versa.

Proof.

Claim: $\displaystyle \mathbb {M} ( \xi ) \subseteq \mathbb {M} (U)$

Now, pick an element $\displaystyle [a, \infty ) \in \xi$,
and write $\displaystyle [a, \infty ) = \bigcup _{r>0} [a,a+r) = \bigcup _{r>0} \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r )$

Note that $\displaystyle (a- \frac {1}{n} ,a+r) \in U \subset \mathbb {M}(U) \ \ \ \ \ \forall n \in \mathbb {N}$, so $\displaystyle \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r ) \in \mathbb {M}(U)$.

Therefore $\displaystyle \bigcup _{r>0} \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r ) \in \mathbb {M}(U)$

So I have $\displaystyle \xi \in \mathbb {M}(U)$, which implies that $\displaystyle \mathbb {M} ( \xi ) \subset \mathbb {M} (U)$

On the other hand, pick $\displaystyle (a,b) \in U$, then $\displaystyle (a,b) = \bigcap _{a<r<b} \bigcup _{n=1}^{ \infty } [r+ \frac {1}{n} , \infty ) \in \mathbb {M} ( \xi )$ for the same reason as above.

Therefore I have $\displaystyle \mathbb {M}(U) \subset \mathbb {M}( \xi )$.

Is this correct? Thank you.

2. Hello,

For the first part, it would have been shorter to say that $\displaystyle [a,\infty)$ is the complement of $\displaystyle (-\infty,a)$, which can be written as $\displaystyle \bigcup_{n> 0} (a+n,a)$, which belongs to $\displaystyle \mathbb{M}(U)$
But what you did is very correct too !

On the other hand, pick $\displaystyle (a,b) \in U$, then $\displaystyle (a,b) = \bigcap _{a<r<b} \bigcup _{n=1}^{ \infty } [r+ \frac {1}{n} , \infty ) \in \mathbb {M} ( \xi )$ for the same reason as above.
Hmm no.
Because the intersection is equal to $\displaystyle (b,\infty)$ or something like that.

Let :
$\displaystyle A_n=\left[a-\frac 1n,\infty\right)$ and $\displaystyle B=[b,\infty)$

Consider $\displaystyle C=\left(\bigcup_{n>0} A_n\right)\cap B^c$

Can you finish it ?

Anyway, good work you've done !