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Let $I\subseteq\mathbb{R}$ be a closed interval, and let $f:I\rightarrow\mathbb{R}$ be a function which is such that for every closed subinterval $D\subseteq I$, $f$ satisfies the intermediate value theorem on $D$.

This does not guarantee continuity of $f$, as we can see with $\sin(1/x)$ on any closed interval containing $0$.

Question: How badly discontinuous can $f$ be? Is there a sense in which $f$ is "close" to be continuous?

Let $X\subseteq I$ be the set of points where $f$ is not continuous. Is $X$ small in some sense? Perhaps $X$ is finite, or $X$ has measure $0$?

semisimpleton
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To convert my comment to an answer: Functions of one real variable satisfying the "intermediate value property" are called Darboux functions. Given a function $f: {\mathbb R}\to {\mathbb R}$, let $C_f$ denote the set of continuity points of $f$. It is a classical result (and a nice exercise) that a subset $E\subset {\mathbb R}$ is a $G_\delta$-subset (i.e. the intersection of countably many open subsets of ${\mathbb R}$) if and only if there exists a function $f: {\mathbb R}\to {\mathbb R}$ such that $E=C_f$. It turns out (see references [a4], [a16] here) that the same holds if one restricts to Darboux functions. In other words, discontinuity sets of Darboux functions are the same as discontinuity sets of general functions of one variable. In particular, both of your conjectures about discontinuity sets of Darboux functions are quite wrong.

Moishe Kohan
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