# Why Continuity set is a borel set?

$\def\R{\mathbb R}$Let $A= \{x: f \text{ is continuous at$x$}\}$ for $f : \R\to \R$ , why is $A$ Borel measurable?

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Think about $\liminf$ and $\limsup$ – M Turgeon Sep 18 '12 at 19:11

The set $A$ is what is called a $G_\delta$ set--a countable intersection of open sets. Since open sets are Borel measurable, and since countable intersections of measurable sets are measurable, then $A$ is measurable.
Let $A_{a,b}=\{x\mid \exists\delta>0\colon\forall y\in (x-\delta,x+\delta)\colon a<f(y)<b\}$.
Then $A_{a,b}$ is open (why?) and $$A=\bigcap_{n\in \mathbb N}\bigcup_{q\in \mathbb Q}A_{q,q+\frac1n}$$ (again: why?) is a countable intersection of open sets.