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This may be obvious, but if we consider the ordinary Cantor set $C$ (removing middle thirds), why is the characteristic function $\chi_C$ continuous $\lambda$-a.e.? The explanation in my book is that the Cantor-set is compact and nowhere dense, but somehow I do not see why this implies that $\chi_C$ is discontinuous on $C$ and continuous on $[0,1] \setminus C$

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    Hint: $\mathbb R \setminus C$ is an open set, and $\chi_C$ is constont on that open set. – 2017-02-02
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    The "nowhere dense" part is used to prove that $\chi_C$ is discontinuous on $C$. The continuity on the complement instead follows because $C$ is closed. – 2017-02-02
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    @Del How would I exactly prove that $\chi_C$ is discontinuous on $C$? (if it is not too long) – 2017-02-02
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    Nowhere dense implies that for every point $x\in C$, for every open interval $I$ containing $x$, there is a point $y\in I$ which is not in $C$. Therefore there are points arbitrarily close to $x$ on which $\chi_C$ is $0$, while on $x$ it clearly is $1$. – 2017-02-02
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    @Del Thanks a lot. – 2017-02-02
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    You're welcome. Also: there exist compact nowhere dense sets with positive measure, therefore to obtain continuity a.e. you really need that $C$ be of measure zero. – 2017-02-02
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    @Del Ah yes, I think of the construction of the Cantor-like sets $A$ where $\lambda(A) = \alpha$ for any $0 < \alpha < 1$. Thanks again. – 2017-02-02

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