I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 2 on p.38 in Exercises 2B in this book.
Verifiy both bullet points in Example 2.28.
2.28 Example smallest $\sigma$-algebra
- Suppose $X$ is a set and $\mathcal{A}$ is the set of subsets of $X$ that consist of exactly one element: $$\mathcal{A}=\{\{x\}:x\in X\}.$$ Then the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $X\setminus E$ is countable, as you should verify.
- Suppose $\mathcal{A}=\{(0,1),(0,\infty)\}.$ Then the smallest $\sigma$-algebra on $\mathbb{R}$ containing $\mathcal{A}$ is $\{\emptyset,(0,1),(0,\infty),(-\infty,0]\cup [1,\infty),(-\infty,0],[1,\infty),(-\infty,1),\mathbb{R}\},$ as you should verify.
My question is about the second bullet point.
My solution:
$A_1:=\emptyset.$
$A_2:=(0,1).$
$A_3:=(0,\infty).$
$A_4:=(-\infty,0]\cup [1,\infty).$
$A_5:=(-\infty,0].$
$A_6:=[1,\infty).$
$A_7:=(-\infty,1).$
$A_8:=\mathbb{R}.$
Then, $\mathcal{A}=\{A_2,A_3\}.$
$\mathcal{S}:=\{A_1,A_2,A_3,A_4,A_5,A_6,A_7,A_8\}.$
Suppose $\mathcal{T}$ is the smallest $\sigma$-algebra on $\mathbb{R}$ containing $\mathcal{A}$.
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_1\in\mathcal{T}$.
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_8=\mathbb{R}\setminus A_1\in\mathcal{T}$.
Since $\mathcal{T}$ contains $\mathcal{A}$, $A_2\in\mathcal{T}.$
Since $\mathcal{T}$ contains $\mathcal{A}$, $A_3\in\mathcal{T}.$
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_4=\mathbb{R}\setminus A_2\in\mathcal{T}.$
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_5=\mathbb{R}\setminus A_3\in\mathcal{T}.$
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_7=A_2\cup A_5\in\mathcal{T}.$
Since $\mathcal{T}$ is a $\sigma$-algebra, $A_6=\mathbb{R}\setminus A_7\in\mathcal{T}.$
Therefore, $\mathcal{S}\subset\mathcal{T}.$
We prove $\mathcal{S}$ is a $\sigma$-algebra.
(1) $\emptyset=A_1\in\mathcal{S}.$
(2)
$A_8=\mathbb{R}\setminus A_1\in\mathcal{S}.$
$A_4=\mathbb{R}\setminus A_2\in\mathcal{S}.$
$A_5=\mathbb{R}\setminus A_3\in\mathcal{S}.$
$A_2=\mathbb{R}\setminus A_4\in\mathcal{S}.$
$A_3=\mathbb{R}\setminus A_5\in\mathcal{S}.$
$A_7=\mathbb{R}\setminus A_6\in\mathcal{S}.$
$A_6=\mathbb{R}\setminus A_7\in\mathcal{S}.$
$A_1=\mathbb{R}\setminus A_8\in\mathcal{S}.$
(3)
$A_1\cup A_1=A_1\in\mathcal{S}.$
$A_2\cup A_2=A_2\in\mathcal{S}.$
$A_3\cup A_3=A_3\in\mathcal{S}.$
$A_4\cup A_4=A_4\in\mathcal{S}.$
$A_5\cup A_5=A_5\in\mathcal{S}.$
$A_6\cup A_6=A_6\in\mathcal{S}.$
$A_7\cup A_7=A_7\in\mathcal{S}.$
$A_8\cup A_8=A_8\in\mathcal{S}.$Since $A_1\subset A_2$, $A_1\cup A_2=A_2\in\mathcal{S}.$
Since $A_1\subset A_3$, $A_1\cup A_3=A_3\in\mathcal{S}.$
Since $A_1\subset A_4$, $A_1\cup A_4=A_4\in\mathcal{S}.$
Since $A_1\subset A_5$, $A_1\cup A_5=A_5\in\mathcal{S}.$
Since $A_1\subset A_6$, $A_1\cup A_6=A_6\in\mathcal{S}.$
Since $A_1\subset A_7$, $A_1\cup A_7=A_7\in\mathcal{S}.$
Since $A_1\subset A_8$, $A_1\cup A_8=A_8\in\mathcal{S}.$Since $A_2\subset A_8$, $A_2\cup A_8=A_8\in\mathcal{S}.$
Since $A_3\subset A_8$, $A_3\cup A_8=A_8\in\mathcal{S}.$
Since $A_4\subset A_8$, $A_4\cup A_8=A_8\in\mathcal{S}.$
Since $A_5\subset A_8$, $A_5\cup A_8=A_8\in\mathcal{S}.$
Since $A_6\subset A_8$, $A_6\cup A_8=A_8\in\mathcal{S}.$
Since $A_7\subset A_8$, $A_7\cup A_8=A_8\in\mathcal{S}.$Since $A_2\subset A_3$, $A_2\cup A_3=A_3\in\mathcal{S}.$
Since $A_4=\mathbb{R}\setminus A_2$, $A_2\cup A_4=A_8\in\mathcal{S}.$
$A_2\cup A_5=A_7\in\mathcal{S}.$
$A_2\cup A_6=A_3\in\mathcal{S}.$
Since $A_2\subset A_7$, $A_2\cup A_7=A_7\in\mathcal{S}.$Since $\mathbb{R}=A_3\cup A_5\subset A_3\cup A_4$, $A_3\cup A_4=A_8\in\mathcal{S}.$
Since $A_5=\mathbb{R}\setminus A_3$, $A_3\cup A_5=A_8\in\mathcal{S}.$
Since $A_6\subset A_3$, $A_3\cup A_6=A_3\in\mathcal{S}.$
$A_3\cup A_7=A_8\in\mathcal{S}.$Since $A_5\subset A_4$, $A_4\cup A_5=A_4\in\mathcal{S}.$
Since $A_6\subset A_4$, $A_4\cup A_6=A_4\in\mathcal{S}.$
$A_4\cup A_7=A_8\in\mathcal{S}.$$A_5\cup A_6=A_4\in\mathcal{S}.$
Since $A_5\subset A_7$, $A_5\cup A_7=A_7\in\mathcal{S}.$Since $A_7=\mathbb{R}\setminus A_6$, $A_6\cup A_7=A_8\in\mathcal{S}.$
So, $\mathcal{S}=\mathcal{T}.$
My question is here:
Suppose $X$ is a set and $\mathcal{A}$ is a finite set of subsets of $X$.
If we do the same as the solution above, can we get the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$?
Best Answer
Yes. The smallest $\sigma$-algebra containing $\mathcal{A}$ is by definition the $\sigma$-algebra generated by $\mathcal{A}$. So taking all the complements and countable unions of elements of $\mathcal{A}$ will produce the smallest $\sigma$-algebra containing $\mathcal{A}$