# Borel sets

• Mar 8th 2010, 05:02 PM
harriette
Borel sets
How do I prove the following?

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

Thank you!
• Mar 8th 2010, 05:53 PM
Focus
Quote:

Originally Posted by harriette
How do I prove the following?

If X is a Borel set, and $a \in \mathbb{R}$, prove that $X+a=\{x+a, x \in X\}$ 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, 05:02 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.

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 $\cup(A_i+a)=\cup(A_i)+a$, 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.
In other words use the Monotone Class theorem, also known as Dynkin's lemma or the $\pi-\lambda$ theorem.