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user3078439
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If $f : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous, why are $f^-$ and $f^+$ measurable?
Thank you so much for this detailed answer. There is something that seems strange to me: if $f$ being Scott-continuous truly means that for all $x \leq y$, $f(x) \supseteq f(y)$, then in particular $f(-a) \supseteq f([-a,a])$? So this function can't afford to be very "thin" (if we think of it as an approximation to a real function, as is the goal of the paper).
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