2
$\begingroup$

Now, I have done some work on this and by using the definition of the derivative, I come to the conclusion that if such a set $A$ exists, then I should have \begin{align*} \lim\limits_{x\rightarrow0^{+}}\frac{m(A\cap(0,x))}{x} = \frac{1}{2} = \lim\limits_{x\rightarrow0^{-}}\frac{-m(A\cap(x,0))}{x} \end{align*} But now, I tried to create such an $A$ in the following way: $A = \bigcup \limits_{n=1}^{\infty} (\frac{1}{2^n}, \frac{3}{2^{n+1}})$. The idea was, that if I took an interval of measure $\frac{1}{2^{n+1}}$ in each interval $(\frac{1}{2^n}, \frac{1}{2^{n+1}})$, then eventually I could have 'half' of $(0,x)$ in each $A\cap(0,x)$ set, as the limit requires. Note that I'm working only for the part $x>0$, if I manage this I think that by symmetry it will work out in the $x<0$ part too. But I cannot prove that for this $A$ the limit is indeed $\frac{1}{2}$, so I'm not sure I'm on the right path.

$\endgroup$
1
  • $\begingroup$ When I tried to nail this down it seemed to me that intervals near zero had to be cut into more and more parts, in order to get convergence. $\endgroup$
    – coffeemath
    Commented Oct 6, 2013 at 20:13

2 Answers 2

2
$\begingroup$

Your idea can be made to work, though it gets a bit messy. If $0<x<1$, there is a unique $n(x)\in\Bbb N$ such that $$\frac1{2^{n(x)+1}}\le x<\frac1{2^{n(x)}}\;.$$

For $m\in\Bbb N$ let $I_m=\left[\frac1{2^{m+1}},\frac1{2^m}\right)$, and let $A_m=\{x\in I_m:\lfloor 4^{m+1}x\rfloor\text{ is even}\}$. Let $A^+=\bigcup_{m\ge 0}A_m$; since $2^{n(x)+1}\le 4^{n(x)+1}x<2^{n(x)+2}$ for $x\in(0,1)$, it’s not hard to see that $A_m$ is a finite union of intervals and that $m(A_m)=\frac12m(I_m)$ for each $m\in\Bbb N$ and hence that $m(A^+)=\frac12$.

For $x\in(0,1)$ let $$f(x)=\frac{m\big(A^+\cap(0,x)\big)}x\;.$$

Let $x\in I_m$, and let $k=\lfloor 4^{m+1}x\rfloor$; $2^{m+1}\le k<2^{m+2}$, and $x\in A^+$ iff $k$ is even. Thus, on the interval $$\left[\frac{k}{4^{m+1}},\frac{k+1}{4^{m+1}}\right)$$ the function $f$ is increasing if $k$ is even and decreasing if $k$ is odd.

It’s not hard to check that if $k$ is even, then

$$\begin{align*} \frac12&=f\left(\frac{k}{4^{m+1}}\right)\\\\ &<f\left(\frac{k+1}{4^{m+1}}\right)\\\\ &=\frac{\frac{k}{2\cdot4^{m+1}}+\frac1{4^{m+1}}}{\frac{k+1}{4^{m+1}}}\\\\ &=\frac{k+2}{2k+2}\\\\ &=\frac12+\frac1{2k+2}\\\\ &\le\frac12+\frac1{2^{m+2}+2}\;, \end{align*}$$

while if $k$ is odd, then

$$\begin{align*} \frac12&=f\left(\frac{k+1}{4^{m+1}}\right)\\\\ &<f\left(\frac{k}{4^{m+1}}\right)\\\\ &=\frac{\frac{k+1}{2\cdot4^{m+1}}}{\frac{k}{4^{m+1}}}\\\\ &=\frac{k+1}{2k}\\\\ &=\frac12+\frac1k\\\\ &<\frac12+\frac1{2^{m+1}}\;. \end{align*}$$

It follows that $\lim\limits_{x\to 0^+}f(x)=\frac12$ and hence that if $A^-=\{-x:x\in A^+\}$, then the set $A=A^-\cup A^+$ has the desired property.

$\endgroup$
2
  • $\begingroup$ My only question is, why is $f$ increasing when $k$ is even? As far as I can see, for $x_1, x_2$ with $\frac{k}{4^{m+1}} \leq x_1 < x_2 < \frac{k+1}{4^{m+1}}$, the value of $f$ on $x_1$ is equal to $\frac{m(A^{+}\cap (0,x_1))}{x_1}$ which is strictly lower than the value $\frac{m(A^{+}\cap (0,x_1))}{x_2}$ which is also strictly lower than $\frac{m(A^{+}\cap (0,x_1))}{x_2} + \frac{m(A^{+} \cap (x_1,x_2)}{x_2} = f(x_2)$. This happens because at the interval $[\frac{k}{4^{m+1}}, \frac{k+1}{4^{m+1}})$, all numbers belong to $A^{+}$. $\endgroup$
    – John Doe
    Commented Oct 7, 2013 at 20:47
  • 1
    $\begingroup$ @John: $f$ increases on intervals in $A^+$ because if $0<a<b$ and $c>0$, then $$\frac{a}b<\frac{a+c}{b+c}\;.$$ Think of $\frac{a}b$ as the value of $f(x)$, and $\frac{a+c}{b+c}$ as the value of $f(x+c)$. $\endgroup$ Commented Oct 7, 2013 at 20:57
2
$\begingroup$

Divide the interval $[2^{-k-1}, 2^{-k}]$ into $N(k)$ subintervals which are alternately included in $A$ or excluded from it. Then the difference quotient $(f(x) - f(0))/x$ for $x$ in the first subinterval will vary between $1/2$ and $\frac{1/2 + 1/N(k)}{1 + 1/N(k)}$. So as long as $N(k) \rightarrow \infty$ the limit will exist at $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .