1. ## Sigma-algebra

Hello,

I have been given the follwoing subsets of $\displaystyle \mathbb{R}$: $\displaystyle A=\left[0,1 \right]$, $\displaystyle B=\left]\tfrac{1}{2},\infty \right[$ and $\displaystyle C= \mathbb{Z}$.

I wish to show that the sets $\displaystyle \left\{-1,-2,-3,... \right\}$, $\displaystyle \left\{0\right\}$ and $\displaystyle A=\left[0,\tfrac{1}{2}\right] \cup \left\{1\right\}$ belong to the $\displaystyle \sigma$-algebra $\displaystyle \sigma(\left\{A,B,C \right\})$.

Would it be enough to, tediously, write out the $\displaystyle \sigma$-algebra and check that the subsets are indeed included or is there an easier method?

Thanks.

2. ## Re: Sigma-algebra

(you are using $A$ to denote two different sets)

Is this an example of what you are looking for? (not tedious)

using $A=[0,1]$

$\displaystyle B^c\cap A\cap C=\{0\}$

3. ## Re: Sigma-algebra

Yes, I see. That was a typo.

I wish to show that the sets $\displaystyle \left\{-1,-2,-3,... \right\}$, $\displaystyle \left\{ 0 \right\}$ and $\displaystyle \left[0,\tfrac{1}{2} \right] \cup \left\{1 \right\}$ belong to $\displaystyle \sigma(\left\{A,B,C \right\})$.

The $\displaystyle \sigma$-algebra contains sets which are obtained by countable unions, intersections and complements of the sets $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$.
I did an example with two sets only and thought that it was a bit tedious and therefore I wanted to know if there is a quicker method of constructing a $\displaystyle \sigma$-algebra.

This is what I think:

By definition $\displaystyle B^{c} \in \sigma(\left\{A,B,C \right\})$.
By definition $\displaystyle A \in \sigma(\left\{A,B,C \right\})$.
By definition $\displaystyle C\in \sigma(\left\{A,B,C \right\})$.

By the properties of a $\displaystyle \sigma$-algebra we have that $\displaystyle B^{c} \cap A \cap C = \left\{ 0 \right\} \in \sigma(\left\{A,B,C \right\})$.

Using the same method as above we see that $\displaystyle B^{c} \cap A^{c} \cap C = \left\{ -1,-2,-3,... \right\} \in \sigma(\left\{A,B,C \right\})$. Correct?

Correct