Prove that the set $A=\{x\in\mathbb{R}:f' \text{ exists}\}$ is Borel whenever $f$ is continuous

Asked Jul 26 '20 at 16:02

Active Jul 26 '20 at 16:02

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Question: Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous. Prove that the set of points where the derivative of $f$ exists is Borel measurable.

My Thoughts: So, we are really trying to prove that the set $A=\{x\in\mathbb{R}:f' \text{ exists}\}$ is open whenever $f$ is continuous. So, if we can show that $A$ is open, then we would be done. But, I am not sure if I should just play with the definition of differentiability, or if there is something topological I should be considering.

Any help, suggestions, ideas, etc. are greatly appreciated! Thank you.

asked Jul 26 '20 at 16:02

User7238

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A Borel measurable set is not necessary open. The Borel sigma-algebra is just generated by the open sets. – Mark Jul 26 '20 at 16:03
Ah, right, so I should be trying more for the countable union of closed or open set tactic? – User7238 Jul 26 '20 at 16:06
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Maybe you can take a look at this question: https://math.stackexchange.com/questions/105810/continuous-functions-are-differentiable-on-a-measurable-set – Mark Jul 26 '20 at 16:08
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See also Characterization of sets of differentiability. – Dave L. Renfro Jul 26 '20 at 16:31

Prove that the set $A=\{x\in\mathbb{R}:f' \text{ exists}\}$ is Borel whenever $f$ is continuous

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