I fail to see how this Borel set is an element of the Borel $\sigma$-algebra.

The Borel Set:

$$ \{1\} = \bigcap_{n=1}^{\infty} (1-1/n,1] = \mathbb{R}\setminus \left(\bigcup_{n=1}^{\infty} \mathbb{R}\setminus (1-1/n,1]\right) $$

I've attempted to use these definitions to see if the Borel set is an element of the Borel $\sigma$-algebra but i haven't had any luck :(

Defintion: A family $\mathcal F$ of subsets of $\Omega$ is said to be a $\sigma$-algebra on $\Omega$ if:

(A.1) $\Omega\in\mathcal F$

(A.2) $\ A\in\mathcal F\implies\ A^c\in\mathcal F$

(A.3) $\ A_1,A_2,...\in\mathcal F\implies\bigcup _{n=1}^\infty A_n\in\mathcal F$


The Borel $\sigma$-algebra on $\mathbb R$ is the $\sigma$-algebra B($\mathbb R$) generated by the $\pi$-system $\mathcal J$ of intervals (a, b], where a < b in $\mathbb R$ (We also allow the possibility that a = $-\infty$ or b =$\infty$) Its elements are called Borel sets. For A $\in$ B($\mathbb R$), the $\sigma$-algebra

B(A)= {B $\subseteq$ A: B $\in$B($\mathbb R$)} of Borel subsets of A is termed the Borel $\sigma$-algebra on A.

  • 5
    $\begingroup$ I think what you are supposed to show here is that the set $\{1\}$ is an element of the Borel $\sigma$-algebra. A single set is not a $\sigma$-algebra! $\endgroup$
    – PhoemueX
    Dec 30, 2014 at 19:36
  • $\begingroup$ Why would you ask whether $\{1\}$ is a $\sigma$-algebra? Who or what gave you the idea that it might be? $\endgroup$
    – David K
    Dec 30, 2014 at 19:42
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    $\begingroup$ @David K Excuse me, that is none of your concern, edits have been made. I obviously have a few misconceptions and blanks in my understanding of the subject. That is why I am here. $\endgroup$
    – Brofessor
    Dec 30, 2014 at 19:56
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    $\begingroup$ Based on edits it seems you came up with the notion yourself, which is fine. Indeed that is how one learns. If the notion had come from somewhere else, though, it might have some implications on what you need to do (e.g. if it was written in exactly those words on a homework assignment). That is why I asked. $\endgroup$
    – David K
    Dec 30, 2014 at 20:03
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    $\begingroup$ Further, you can generate the Borel-$\sigma$-Algebra on $\mathbb{R}$ with $ \{ \; \{ x \} \; | \; x \in \mathbb{R} \} $. $\endgroup$
    – kummerer94
    Dec 30, 2014 at 22:28

1 Answer 1


According to your definition of the Borel $\sigma$-algebra $\text B(\mathbb R)$, the interval $I_n=(1-1/n,1]$ is contained in $\text B(\mathbb R)$ for every $n$. Thus, according to clause (A.2), so is their complements $I_n^c=(-\infty,1-1/n]\cup(1,\infty).$ Then, according to clause (A.3), you know that the union of the complements $I_n^c$ is also in $\text B(\mathbb R)$.

Hint: at this point, you can use DeMorgan's Law for unions and clause (A.2) again to obtain the desired result.


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