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I am looking to show that if we have a Lebesgue measurable set $E \subseteq \mathbb{R}$ with a density 1 at every element in $E$ and a density of 0 at every element of $\mathbb{R} \backslash E$. Then it must be that $E = \mathbb{R}$ or $E = \emptyset$.

I am working through Axlers book on measure theory and we have defined the density of $E$ at a number $b \in \mathbb{R}$ to be $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t}$.

From the Lebesgue Density Theorem I know that for a Lebesgue measurable set $E \subset \mathbb{R}$, the density of $E$ is 1 at almost every element of $E$ and is 0 at almost every element of $\mathbb{R} \backslash E$. So the difference to this case is that we are saying its true everywhere as opposed to almost everywhere.

So so far I have that:

  • For all $b \in E$ we have $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t} = 1$
  • For all $b \in \mathbb{R} \backslash E$ we have $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t}=0$

Intuitively it makes sense why only $\mathbb{R}$ and $\emptyset$ work but I am having trouble putting it into a complete proof. Any help is greatly appreciated. Thanks in advance!

Charlie Sandberg
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    Just guessing: Can we show that E is open & closed? – DanielWainfleet Nov 02 '20 at 06:20
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    The book mentioned in this question was published in Springer's Graduate Texts in Mathematics series as an Open Access book. Thus it is legally available for free at the following link: http://measure.axler.net/. – Sheldon Axler Nov 02 '20 at 07:26

1 Answers1

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The function defined on $(0, \infty)$ $$f(x) = m((0, x]\cap E)$$ has a derivative everywhere, and $f'$ takes values in $\{0,1\}$, but it also has the Darboux property.

Note: the density is defined with a "symmetric derivative", so in general it may not be equivalent with the usual derivative.

$\bf{Added:}$ While the existence of the symmetric derivative does not imply the existence of the derivative, note that our functions satisfies $$0 \le \frac{f(x+h)- f(x) }{h} \le 1$$ and so $$0\le \frac{f(x+h)- f(x) }{h}\le 2 \cdot \frac{f(x+h)- f(x-h)}{2 h}$$

Therefore, if at some point we have $$\lim_{h\to 0}\frac{f(x+h)- f(x-h)}{2 h} = 0 $$ then $f'(x) = 0 $. Working with the function $x-f(x)$ we also conclude that if the symmetric derivative is $1$ then the derivative is $1$.

Therefore, $f$ has a derivative everywhere on $(0, \infty)$ equal to $0$ or $1$.

orangeskid
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    should be comment, not answer – mathworker21 Nov 09 '20 at 12:16
  • The absolute value function makes it clear that the Darboux property doesn't always hold for the symmetric derivative. – John Dawkins Nov 09 '20 at 16:38
  • @John Dawkins: In this case symmetric derivative $=0$ or $1$ implies that the derivative exists. Added some details. – orangeskid Nov 10 '20 at 01:12
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    FYI, letting $\underline{LD}$ be (arbitrarily uncentered) lower Lebesgue density, $\overline{LD}{\text{sym}}$ be (centered) upper symmetric Lebesgue density, etc. we have: $0 ; \leq ; \underline{LD} ; \leq ; \underline{LD}{\text{sym}} ; \leq ; \overline{LD}{\text{sym}} ; \leq ; \overline{LD} ; \leq ; 1,$ and thus if $\underline{LD} = \overline{LD} = 0$ (i.e. uncentered Lebesgue density exists and equals $0),$ then we have $\underline{LD}{\text{sym}} = \overline{LD}_{\text{sym}} = 0$ (i.e. centered Lebesgue density exists and equals $0),$ (continued) – Dave L. Renfro Nov 10 '20 at 08:59
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    and similarly when $0$ is replaced by $1.$ However, for a given measurable set, this can fail for values other than $0$ and $1,$ although only for a fairly small set of points --- the set of points at which this can fail is $\sigma$-upper porous (thus, a bit smaller than being simultaneously measure zero and meager in the Baire category sense) but I also believe the set of points can still have continuum many points in every interval (certainly $\sigma$-upper porous sets can have continuum many points in every interval, (continued) – Dave L. Renfro Nov 10 '20 at 09:00
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    but here I'm talking about the set of points where the symmetric density exists and the ordinary density doesn't exist). See Theorem 5 on p. 266 of Symmetric and ordinary differentiation by Belna/Evans/Humke (1978). Also related to symmetric vs. ordinary Lebesgue density, see this answer and this question and answer and this 31 July 2009 sci.math post. – Dave L. Renfro Nov 10 '20 at 09:00
  • @Dave L.Renfro: Thank you for the interesting info. For this particular problem we use the easy inequality $\bar{LD} \le 2 \bar{LD}_{\textrm{sym}}$, really simple analysis. The general one should be delicate, even the Lebesgue theorem is. – orangeskid Nov 10 '20 at 09:15
  • @Dave L.Renfro: I wonder if it is possible that the usual derivative exists everywhere except the case when the set ( or the complement) has measure $0$. I think not but I don't have an argument now. – orangeskid Nov 10 '20 at 09:46
  • I'm not sure what you're referring to regarding "usual derivative exists everywhere except", but my comments here give a brief historical survey of results known about the set of points where an everywhere symmetrically differentiable function fails to have an ordinary derivative. See also my answer to Existence of the derivative at a point is implied by a version of the symmetric derivative plus continuity. – Dave L. Renfro Nov 10 '20 at 11:29
  • @orangeskip What I was trying to get at was that f(x) = |x| has a symmetric derivative at each point but does not have the Darboux property. – John Dawkins Nov 12 '20 at 00:11
  • @John Dawkins: Fair point, yes indeed :-) – orangeskid Nov 12 '20 at 00:50