If $f$ is continuous, is it (Lebesgue) measurable ? Namely, $\{x\ :\ f(x)<\alpha\}\in\mathcal M\ \forall\alpha$ ?
If not what are some counter examples?
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@John11 This question does not have a satisfying answer. – Jean-Claude Arbaut Jun 03 '18 at 09:49
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2See here for a counterexample https://math.stackexchange.com/questions/479441/example-of-a-continuous-function-that-is-not-lebesgue-measurable?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – Severin Schraven Jun 03 '18 at 09:53
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1In standard terminology $f:\mathbb R \to \mathbb R$ is called Lebesgue measurable if the inverse image of every Borel set is Lebesgue measurable. With this definition the answer is surely 'yes'. – Kavi Rama Murthy Jun 03 '18 at 12:18
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If the codomain $\sigma$-algebra considered is the Borel $\sigma$-algebra:
Yes it is Lebesgue measurable if continuous. You can show first that an upper (lower) semicontinuous function is measurable and then use the fact that a function that is both lower semicontinuous at upper semicontinuous at a point $x_0$ is continuous at that point.
In order to show that an upper (lower) semicontinuous function $f$ is measurable, show that $f^{-1}(]-\infty ,\alpha [)$ are closed and then the result would follow from the fact that closed sets are measurable.
John11
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2The Lebesgue sigma algebra is not generated by open sets (this would be the Borel sigma algebra). – Severin Schraven Jun 03 '18 at 09:55
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@SeverinSchraven Thank you for the remark. I corrected my answer to specify when the proof works. – John11 Jun 03 '18 at 10:07