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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 $[0,1]$

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  • $\begingroup$ 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
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$\sigma^X(\{I\})$ , 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 $\sigma^X \{A_1, A_2\} = \sigma^X \{A_1^\complement, A_2^\complement\}$ ...

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

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