sigma algebra over R

**Maccaman** · August 7th 2009, 04:44 PM

Consider a $\sigma$ -algebra $\mathcal{F}$ on $\mathbb{R}$ containing all intervals $(-\infty, x], x \in \mathbb{R}$ . Show that it must contain the sets of the form $(a,b], (a,b)$ and $\lbrace a \rbrace$ $($ where $a < b),$ 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 $\mathcal{F}$ is closed under complements and under countably infinite solutions.

**Moo** · August 8th 2009, 12:08 AM

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

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

Intersect $(x,\infty)$ with U to get $(x,y)$ (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 $\{b\}$
So for any x in $\mathbb{R}$ , $\{x\}$ 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.

**awkward** · August 8th 2009, 06:26 AM

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 $y<x$
$(-\infty,y]\cap (-\infty,x]=[y,x]$
[snip]

Ahem (clears throat).

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

**Moo** · August 8th 2009, 06:40 AM

Originally Posted by awkward

Ahem (clears throat).

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

Oh sorry !

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

**Maccaman** · August 8th 2009, 03:35 PM

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

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