Let $X=[0,1]$. Describe the sigma algebra generated by these sets.

$I=(0,\frac{1}{2})$. Define the $\sigma(\{I\})$ to denote such $\sigma$-algebra.

I am having problems on how to see what elements are in the $\sigma$-algebra. I am assuming that they are intervals $(a,b) \subseteq I$ that generates the $\sigma$-algebra. For instance, $I \in \sigma(\{I\})$ and $\emptyset \in \sigma(\{I\})$, since $\emptyset = I\cap \emptyset$. Now, unions should belong to the $\sigma(\{I\})$, that is, $(a_1,b_1) \cup (a_2,b_2) \in \sigma(\{I\})$ provided $(a_i,b_i) \subset I$ for $i=1,2$ and in particular it can be extended easily to countable unions. Now, let $(a,b)\in \sigma(\{I\})$, then the complement should be also in $\sigma(\{I\})$. Thus, $$(0,a]\cup[b,\frac{1}{2}) \in \sigma(\{I\}).$$ So that intervals of the form $(a,b]$ and $[a,b)$ are in $\sigma(\{I\})$. Finally, the closed interval $[a,b] \subset I$ is in the $\sigma(\{I\})$ because $$[a,b]= \bigcap_{n\in \mathbb{N}}(a-\frac{1}{n},b+\frac{1}{n}).$$ So that we have: $$\sigma(\{I\}) = \{\emptyset, I, (a,b],[a,b),[a,b]\}.$$ Does the singletons $\{0\}, \{1/2\}$ also belongs to the set? Also, it is not enough to define the sigma algebra as follows: $$\sigma(\{I\}) := \{\mathcal{O}\cap I : \mathcal{O} \subseteq \mathcal{B}_X\},$$ where $\mathcal{B}_X$ is the sigma algebra of Borel sets on $X=[0,1]$?

$A_1 = [0,\frac{1}{4}), A_2 = (\frac{3}{4},1]$. Find $\sigma(\{A_1,A_2\})$.

We first note that $A_1 \cap A_2 = \emptyset \in \sigma(\{A_1,A_2\})$. Also, their union $A_1 \cup A_2 \in \sigma(\{A_1,A_2\})$. Now, for $[a,b) \subseteq A_1$, we have $[a,b)^c = [0,a)\cup[b,\frac{1}{4})$. In particular, $(a,b) = [0,b)\cap(a,\frac{1}{4}]$, so that $(a,b)$ is also in the $\sigma$-algebra. As before $[a,b] = \bigcap_{n\in \mathbb{N}}(a-\frac{1}{n},b+\frac{1}{n})$. Thus, $$\sigma(\{A_1,A_2\}) = \{\emptyset, A_1,A_2, (a,b],[a,b),[a,b]\}.$$ Also, are the singletons $\{1/4\}, \{3/4\}$ in the $\sigma$-algebra. I was wondering if it is enough to define: $$\sigma(\{A_1,A_2\}) := \{\mathcal{O}\cap (A_1 \cup A_2) : \mathcal{O} \subseteq \mathcal{B}_X\},$$ where $\mathcal{B}_X$ is the sigma algebra of Borel sets on $X=[0,1]$?

$A_1 = [0,\frac{3}{4}), A_2 = (\frac{1}{4},1]$. The sigma algebra of these sets would be the sigma algebra generated by $X = [0,1]$, since $A_1 \cup A_2 = X$, which I could guess is the Borel sigma algebra of subsets of $[0,1]$

  • $\begingroup$ The $\sigma$-algebra generated by $\{(0,1/2)\}$ is $\{\emptyset,[0,1],(0,1/2), {0}\cup\{[1/2,1]\}\}.$ You can do the others like this one. $\endgroup$ – Filburt Feb 15 '17 at 23:03
  • $\begingroup$ Let see If i get the concept. $I^c = 0 \cup [1/2,1]$. and basically this yields to $[0,1] = 0 \cup [1/2,1] \cup(0,1/2)$. Also, since we are considering $X=[0,1]$, then $I^c = X \sim I$ right? $\endgroup$ – richitesenpai Feb 15 '17 at 23:18
  • $\begingroup$ yeap, you got it! (use \backslash) $\endgroup$ – Filburt Feb 15 '17 at 23:19
  • $\begingroup$ Thanks you. I think the exercise description in the homework was not very clear. $\endgroup$ – richitesenpai Feb 15 '17 at 23:19

$\sigma^X(\{I\})$ , the sigma algebra over $X$ of $\{I\}$, is its closure under countable union, countable intersections, and complements relative to $X$.

Long story short, that is just $\{\emptyset, I, X\setminus I, X\}$, or $\{\emptyset, I, I^\complement, X\}$ where $I^\complement$ is taken as the complement relative to $X$.

Which is $\{\emptyset, (0;\tfrac 12), \{0\}\cup[\tfrac 12;1], [0;1]\}$.

$\sigma^X(\{A_1,A_2\})$ is however, slightly more involved.

Now notice that $[0;\tfrac 34),(\tfrac 34;1]$ are disjoint intervals, and as such it simplifies to: $$\sigma^X\{A_1,A_2\}=\{\emptyset, A_1,A_1^\complement \cap A_2^\complement, A_2, A_1^\complement, A_1\cup A_2, A_2^\complement, X\}$$

$$\sigma^{[0;1]}(\{[0;\tfrac 34),(\tfrac 34;1]\} ~=~\{\emptyset, [0;\tfrac 14),\underline{\qquad},\underline{\qquad}, \underline{\qquad},\underline{\qquad\qquad}, \underline{\qquad}, [0;1]\}$$

Now, $[0;\tfrac 34 ),(\tfrac 14 ;1]$ overlap, but their complements are disjoint.

So as $\sigma^X \{A_1, A_2\} = \sigma^X \{A_1^\complement, A_2^\complement\}$ ...

  • $\begingroup$ Thank you very much. This helps a lot. $\endgroup$ – richitesenpai Feb 16 '17 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.