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.