The question I am attempting:

Let $\Omega = \Bbb R$ and $I_k = [k-1,k)$ for $k \geq 1$ and let $A_n = \sigma\{I_k, k = 1,2,....,n\}$. Show that $\bigcup_{n =1 }^\infty {A_n}$ is not a sigma algebra.

My answer:

In the question it's not specified by $\sigma$ which sigma-algebra they actually mean (perhaps the sigma-algebra consisting of all possible subsets), anyway, by the definition of sigma-algebra,

$I_1 , I_2, ..., I_n \epsilon A_n$ , so in particular ,

I consider $I_k , k \epsilon \Bbb N$. Each such $I_k \epsilon \bigcup_{n =1 }^\infty {A_n}$ but if we consider the countable union i.e. $\bigcup_{k =1 }^\infty {I_k}$ , it does not belong to $\bigcup_{n =1 }^\infty {A_n}$ since there does not exist any $n \geq 1$ such that $\bigcup_{k =1 }^\infty {I_k} \epsilon A_n$.

Is my answer correct? And if there is any flaw please point it out and also provide an answer for the question. Thanks in advance...

  • 1
    $\begingroup$ I think $A_n$ is the $\sigma$-algebra generated by the family of intervals $(I_k)_{1 \le k \le n}$. $\endgroup$ – Randy Randerson Aug 22 '17 at 8:31
  • $\begingroup$ you are correct, only you do need to give some reason for why the union you take is not in $A_n$. The $\sigma$ construction is the smallest sigma algebra containing the set. $\endgroup$ – Ittay Weiss Aug 22 '17 at 8:31
  • $\begingroup$ @RandyRanderson Yes, Probably it's the sigma-algebra generated by the family of given intervals. Anyway, each $I_k \epsilon A_n, 1 \leq k \leq n$, right? Then is the proof alright? $\endgroup$ – reflexive Aug 22 '17 at 8:36
  • $\begingroup$ @IttayWeiss Since given any $n \epsilon \Bbb N$ , $\bigcup_{k=n+1}^\infty {I_k}$ is not contained in $A_n$. Is that enough? $\endgroup$ – reflexive Aug 22 '17 at 8:40

Not quite. You have not explained why $\bigcup_{k=1}^\infty I_k = [0,\infty)$ does not belong to any $A_n$.

Hint :

Given a natural integer $n \ge 1$, find a $\sigma$-algebra $M_n$ such that $[0,\infty) \notin M_n$ and $A_n \subseteq M_n$. It follows that $[0,\infty)$ does not belong to any $A_n$.

By the way, the LaTeX code for $x \in A$ is $ x \in A $ .

  • $\begingroup$ The natural choice for $M_n$ should be $M_n = A_{n+1}$. $\endgroup$ – reflexive Aug 22 '17 at 11:17
  • $\begingroup$ You still haven't explained why $[0,\infty)$ does not belong to $A_{n + 1}$. There is a simple choice for $M_n$. $\endgroup$ – Randy Randerson Aug 22 '17 at 11:21
  • $\begingroup$ Would you please write your choice for $M_n$. And if I can provide detailed argument on $[0,\infty) \notin A_n$ given any $n \geq 1$ is my answer correct? $\endgroup$ – reflexive Aug 22 '17 at 11:27
  • $\begingroup$ To your second question, yes, but the easiest way to do that is with the hint I gave you. To your first question, no, I'm not going to do it for you, but I'll give you another hint. You need a $\sigma$-algebra $M_n$ such that $I_k \in M_n$ for $1 \le k \le n$. Therefore $\bigcup_{k=1}^n I_k = [0,n) \in M_n$. So find a $\sigma$-algebra $M_n$ on $[0,n)$ which contains each $I_k$. Then $A_n \subseteq M_n$ and $[0,\infty) \notin M_n$. $\endgroup$ – Randy Randerson Aug 22 '17 at 11:35

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