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!

Printable View

- Mar 8th 2010, 04:02 PMharrietteBorel 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 PMFocus
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 PMharriette
- Mar 11th 2010, 04:33 PMFocus
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 PMmabruka
In other words use the Monotone Class theorem, also known as Dynkin's lemma or the $\displaystyle \pi-\lambda$ theorem.