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I am trying to prove the following claim, but I don't find a proper way using the theorems we have learned. I hope you can help.

Let $f$ be an increasing function in $[0,1]$ and let $a,b\in\Bbb{R},\textrm{ s.t }a<b$.

Consider the following set $$A=\left\{x;\liminf_{h\to 0+}\frac{f(x+h)-f(x)}{h}<a<b<\limsup_{h\to 0+}\frac{f(x+h)-f(x)}{h}\right\}$$ Show that $m(A)=0$ [as a hint, we are given that $A$ is measurable].

I believe that $\displaystyle \lim_{h\to 0+}\frac{f(x+h)-f(x)}{h}$ points out to look on the right derivative at some point $x$, but I don't know whatsoever this fact helps me.

Thanks!

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Galc127
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  • Hint: Any point $x\in A$ is a point at which $f$ is not differentiable. Moreover, any monotone function on an interval is almost everywhere differentiable. – Marc Aug 25 '16 at 07:22
  • @Marc, so if $f$ is almost everywhere differentiable, I can say that the set $B$ of points where it is not differentiable (and $A\subseteq B$) is of measure zero? If so, I need to prove the statement that monotone function on interval is almost everywhere differentiable? – Galc127 Aug 25 '16 at 07:41
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    Yes, exactly. The set $B$ has measure zero for monotone functions on an interval. Jonas Meyer gives some references to this last statement in an answer to the question: http://math.stackexchange.com/questions/17069/monotonecontinuous-but-not-differentiable. – Marc Aug 25 '16 at 07:43
  • @Marc, thanks for helping! – Galc127 Aug 25 '16 at 07:46

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