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Let $\mu $ be Lebesgue outer measure, does there exist a certain fat Cantor $ C $ such that $\mu (C) > 0$ and for every $ x \in C $ the ratio $\frac {\mu (C\cap [x ,x+1/n])}{\mu (C^c\cap [x , x+1/n])} \to \infty$ as $ n\to\infty $

If such set exists we may answer the interesting question When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail?

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    $\begingroup$ What do you conclude from the Lebesgue density theorem? Perhaps: for almost every point of $C$, $$\lim\frac{\mu(C \cap [x,x+1/n])}{1/n} \to 1\qquad\text{and}\qquad\lim\frac{\mu(C^c \cap [x,x+1/n])}{1/n} \to 0$$ $\endgroup$ – GEdgar Jan 20 '18 at 12:03
  • $\begingroup$ @GEdgar thanks for noticing , the limit in my question should be $\to \infty $ not $0$, i edited that $\endgroup$ – ibnAbu Jan 20 '18 at 12:56
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    $\begingroup$ So the question is: can this 'almost everywhere" thing hold everywhere in $C$? Answer: no. At the left-hand end of one of the removed open intervals, your limit is $0$. $\endgroup$ – GEdgar Jan 20 '18 at 13:19
  • $\begingroup$ @GEdgar Have you looked at the original question, the one linked to above? It seems like such an obvious question I can't believe we don't all know the answer - so far all I can prove is a counterexample must be complicated... $\endgroup$ – David C. Ullrich Jan 20 '18 at 14:58

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