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$\def\R{\mathbb R}$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

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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.

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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.

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Perfect! So do you think it is still the case for A={x:x is differentiable} - I learnt that most of these measurable problems are concerned with finding a lim-sup and liminf but what would it be about differentiables? –  Mathfollower Sep 19 '12 at 2:44

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