# Sigma-Algebra problem

1. Sep 6, 2007

### P3X-018

If we define two sets as

$$A = \left\{ (x,y)\in \mathbb{R} \vert x+y<a \right\}$$
$$B = \left\{ (x,y)\in \mathbb{R} \vert x-y<a \right\}$$

and then define the system of sets,

$$\mathbb{D} = \left\{A\vert a\in \mathbb{R}\right\}\cup\left\{B\vert a\in \mathbb{R}\right\}$$

and then let $\sigma(\mathbb{D})$ be the $\sigma$-algebra generated by the system of set $\mathbb{D}$. To show that the sets

$$C = \left\{ (x,y)\in \mathbb{R} \vert x+y\leq a \right\}$$
$$D = \left\{ (x,y)\in \mathbb{R} \vert x-y\leq a \right\}$$

are elements in $\sigma(\mathbb{D})$ for every $a\in \mathbb{R}$.
Some questions arises. Can the sets A and B be considered as elements in $\mathbb{D}$ and hence in $\sigma(\mathbb{D})$ too, or are they just subsets of $\mathbb{D}$?
Now I want to express C as a union of A and the set $S = \left\{ (x,y)\in \mathbb{R} \vert x+y=a \right\}$, and similar for D. But then I have to express the new set in some way to show that it is contained in $\sigma(\mathbb{D})$. (Maybe there are easier approaches?). I was considering to write the new set S as

$$S = \bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \vert x+y<a+\frac{1}{n} \right\}\cap \left\{ (x,y)\in \mathbb{R} \vert x+y\geq a \right\}$$

Where the last set is the complement of A. But will this set even converge towards S, or will it just be empty?

Last edited: Sep 6, 2007
2. Sep 6, 2007

### EnumaElish

I think A and B are not subsets. {A} and {B} are subsets. But, isn't $\sigma$ defined over subsets of $\mathbb{D}$?

I think S is empty, but $\bar S$, defined with an < (instead of an <) is non-empty. See http://en.wikipedia.org/wiki/Nested_sequence_of_intervals

Last edited: Sep 7, 2007
3. Sep 7, 2007

### P3X-018

So A and B are elements in $\sigma(\mathbb{D})$, right? And yes $\sigma(\mathbb{D})$ is defined as the smallest $\sigma-$algebra containing $\mathbb{D}$.

Hmm... Now you have changed your answer of weather S is empty or not. I thought that maybe it would be non-empty because of the way we (in class) defined an interval of the type (a,b] as

$$(a,b] = \bigcap_{n=1}^{\infty}\left(a,\, b+\frac{1}{n}\right)$$

as a intersection of open intervals.
But hmm... then again maybe this wasn't even a good inspiration, since it seems different now that I type it...

But how would you suggest to write S from the 'building blocks' of $\sigma(\mathbb{D})$?

EDIT:
The question really is just, will the infinity intersection be

$$\bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \,\vert\, x+y<a+\frac{1}{n}\right\} = \left\{ (x,y)\in \mathbb{R} \,\vert\, x+y\leq a\right\}$$

Because that is the set I'm intrested in. And that's why I considered that if (a,b] can be writting in that way, why can't the above equation be true?
By this expression S becomes completly useless, since I can just set C equal to that expression, and C will then consist of an infinite intersection of A sets. But then we're back to my first question, are A and B elements in the $\sigma$-algebra even though it is generated by the union of A and B, so that C is can be considered as an element in the algebra?

Last edited: Sep 7, 2007
4. Sep 7, 2007

### P3X-018

Ok instead of making yet another edit, I'll make a new reply, getting confusing wit so edits now.
I think I was confused about $\mathbb{D}$, since I saw it a union of 2 sets, but it's actually a union of 2 system of sets. So the $\mathbb{D}$ is actually,

$$\mathbb{D} = \left\{ A \subset \mathbb{R}^2 \,\vert\, A\in \{A_a\,\vert\, a\in\mathbb{R}\}\quad\mathrm{or}\quad A\in \{B_a\,\vert\, a\in\mathbb{R}\} \right\}$$

and I think that it is clear from this, that A and B are contained in $\mathbb{D}$ and hence also $\sigma(\mathbb{D})$.

5. Sep 7, 2007

### EnumaElish

I think I agree with
$$\bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R}^2 \,\vert\, x+y<a+1/n\right\} = \left\{ (x,y)\in\mathbb{R}^2\vert x+y\le a\right\}$$

I think I was right the first time, then confused myself with nested intervals.

Last edited: Sep 7, 2007