Hello I was wondering if my proof for the following problem is correct and how to show one of the $O_n$ has finite measure?
Suppose E is a given set, and $O_n$ is the open set $O_n = \{x: \inf_y |x-y|<\frac{1}{n}\}$ where the infimum is taken over all y in E. (a) Show that if E is compact, then $ lim(O_n)= m(E) $
my proof) First, I show that $O_n$ tends to E. Clearly, E is in the intersection of $O_n$ so all that is left to show is $O_n$ is in E. Let x belong to the intersection of $O_n$, then there exists a $y_n$ in $O_n$ such that $|x-y_n|<\frac{1}{n} $. This defines a sequence $y_n$ in E which converges to x, and since E is closed, x is in E. Hence E= intersection $O_n$.
Now if I am able to show that one of the $O_n$ has finite measure, then I could use continuity of measure to conclude that $m(O_n)\to m(E)$
How can I do that and is my proof otherwise correct? Thanks in advance.