# Borel sets

• Mar 8th 2010, 04:02 PM
harriette
Borel sets
How do I prove the following?

If X is a Borel set, and $\displaystyle a \in \mathbb{R}$, prove that $\displaystyle X+a=\{x+a, x \in X\}$ is also a Borel set.

Thank you!
• Mar 8th 2010, 04:53 PM
Focus
Quote:

Originally Posted by harriette
How do I prove the following?

If X is a Borel set, and $\displaystyle a \in \mathbb{R}$, prove that $\displaystyle X+a=\{x+a, x \in X\}$ is also a Borel set.

Thank you!

Fix a, and use a monotone class argument. So for any open set B, consider B+a, which generate X+a. It is rather obvious that all the open sets are of the form B+a; If you give me a set N open, N-a is open, and so N=(N-a)+a. Hence the two sigma algebras coincide.
• Mar 11th 2010, 04:02 PM
harriette
Quote:

Originally Posted by Focus
Fix a, and use a monotone class argument. So for any open set B, consider B+a, which generate X+a. It is rather obvious that all the open sets are of the form B+a; If you give me a set N open, N-a is open, and so N=(N-a)+a. Hence the two sigma algebras coincide.

Thank you for your time!
Could you please tell me what a monotone class argument is?
• Mar 11th 2010, 04:33 PM
Focus
Quote:

Originally Posted by harriette
Thank you for your time!
Could you please tell me what a monotone class argument is?

When you show that the generators (pi systems) are the same. So in this case the Borel sets are generated by the open sets. As $\displaystyle \cup(A_i+a)=\cup(A_i)+a$, the sets of the form X+a are generated by U+a where U is open. If you know that the generators are the same, then you know that the sigma algebra they generate are the same as well.
• Mar 11th 2010, 04:38 PM
mabruka
In other words use the Monotone Class theorem, also known as Dynkin's lemma or the $\displaystyle \pi-\lambda$ theorem.