It's extremely well-known that continuous functions are Borel measurable, but what about piecewise continuous functions? For the Lebesgue measure, I suspect that we'd have a proof as simple as "continuous functions are measurable, piecewise continuous functions are just continuous functions with a few bits missing, singletons have measure 0, throw in some additivity argument, done", but I'm unsure about the more general case.
2026-03-25 04:55:45.1774414545
Does piecewise continuous imply Borel measurable?
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Yes, if you write down the inverse image of a Borel set you get a finite union of Borel sets. The question in the title has nothing to do with any measure.