# Thread: sigma algebra over R

1. ## sigma algebra over R

Consider a $\displaystyle \sigma$-algebra $\displaystyle \mathcal{F}$ on $\displaystyle \mathbb{R}$ containing all intervals $\displaystyle (-\infty, x], x \in \mathbb{R}$. Show that it must contain the sets of the form $\displaystyle (a,b], (a,b)$ and $\displaystyle \lbrace a \rbrace$ $\displaystyle ($ where $\displaystyle a < b),$ and the countable unions thereof.

In class my lecturer said that I should keep de Morgan's laws in mind, as well as the fact that $\displaystyle \mathcal{F}$ is closed under complements and under countably infinite solutions.

2. Hello,

You just have to play with the axioms of a sigma algebra.
By combining the property of complements and the property of countable unions, we get, by de Morgan's law, that a sigma algebra is closed under countable intersections.

The complement in $\displaystyle \mathbb{R}$ of $\displaystyle (-\infty,x]$ is $\displaystyle (x,\infty)$ and belongs to the sigma algebra.
Consider $\displaystyle z>x$
$\displaystyle (x,\infty)\cap (-\infty,z]=(x,z]$
(1) So for z>x, (x,z] belongs to the sigma algebra.

Let's consider, for y>x, $\displaystyle U_n=\left(-\infty,y-\tfrac 1n\right]$
A sigma algebra is closed under countable union.
So $\displaystyle U=\bigcup_{n\geq 1} U_n$ belongs to the sigma algebra.
Thus $\displaystyle U=(-\infty,y)$ belongs to the sigma algebra.

Intersect $\displaystyle (x,\infty)$ with U to get $\displaystyle (x,y)$ (2)

Let's take a<b. So by (1) and (2), (a,b] and (a,b) belong to the sigma algebra.
Intersect (a,b] with the complement of (a,b) and you will get $\displaystyle \{b\}$
So for any x in $\displaystyle \mathbb{R}$, $\displaystyle \{x\}$ belongs to the sigma algebra.

I hope all of this is clear... The letters are quite confusing. I suggest you do it on your own once you've catched the tricks...

Edit : and sorry for the mistakes.

3. Originally Posted by Moo
Hello,

You just have to play with the axioms of a sigma algebra.
By combining the property of complements and the property of countable unions, we get, by de Morgan's law, that a sigma algebra is closed under countable intersections.

Let's take $\displaystyle y<x$
$\displaystyle (-\infty,y]\cap (-\infty,x]=[y,x]$
[snip]
Ahem (clears throat).

$\displaystyle (-\infty,y]\cap (-\infty,x]=\color{red}(-\infty,y]$

4. Originally Posted by awkward
Ahem (clears throat).

$\displaystyle (-\infty,y]\cap (-\infty,x]=\color{red}(-\infty,y]$
Oh sorry !

And this isn't even the only mistake...
I'll correct them, thanks

5. I apologize for posting this in the wrong forum. Thanks for moving Mr Fantastic. Thank-you awkward and Moo for your help. Greatly appreciated