Let G={x | f is continuous at x}
Pick some $ x \in G. $
Then $ \exists \ \delta>0 \ s.t.\ |x-y|<\delta \Rightarrow |f(x)-f(y)| < \frac{1}{2} $. But since f is integer valued, $ |f(x)-f(y)| < \frac{1}{2} \Rightarrow f(x)=f(y) $, so f is constant on $ (x-\delta ,x+\delta) $ which means that f is continuous on $ (x-\frac{\delta}{2} ,x+\frac{\delta}{2}) $
$ \Rightarrow (x-\frac{\delta}{2} ,x+\frac{\delta}{2}) \subset G $
So for each $ x \in G, \ \exists $ an open interval about x which is also contained in $ G \ \Rightarrow \ G $ is open.
$ \Rightarrow $ {x | f is not continuous at x} is closed hence Borel.
--Wardbc 12:33, 24 July 2008 (EDT)
