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Working through some measure theory, I came upon the Lebesgue-like decomposition for monotonic functions. In that context, I've cooked up a singular measure $\nu$ on $[0,1]$ about which I know only that it is regular Borel, continuous, and satisfies $\nu K = 0$ for some $K \subset [0,1]$ of full Lebesgue measure.

I would like to conclude that $K$ contains a subset $K'$ of full Lebesgue measure for which each element $x$ satisfies $$ \nu\bigl((x-1/n,x+1/n)\bigr) = \mathcal O(1/n)~~. $$ This would be sufficient to demonstrate that the function mapping $x \mapsto \nu\bigl([0,x]\bigr)$ is singular, in the sense of having an a.e. defined and vanishing derivative (of course, it is continuous as well by continuity of $\nu$).

I'd appreciate any comments on how to show the above bound, and whether that's a good strategy for approaching the problem of defining a singular function from a singular measure. My other thought was that perhaps there is a standard way to 'remove' density from the complement of $K$ (using continuity) sufficiently to form an open set containing $K$ with zero $\nu$-measure, from which the result follows (and the singular set looks much more like the complement of a Cantor set).

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You do not need that condition. Your measure is already singular continuous, so your mapping must give you a singular continuous function. – Thomas Jan 17 '13 at 11:12
@Thomas Sure, but the condition I gave is equivalent to the mapping having a vanishing derivative at $x$, which for singularity must be true for all $x$ in a set of full measure; it's not clear to me how without some work in that direction, the result could come obviously. – Eugene Shvarts Jan 17 '13 at 18:56
Ah, I see, so you want to show that if a measure is singular, then the cumulative distribution function is singular, right? – Thomas Jan 17 '13 at 20:49
@Thomas Spot-on. After doing some more reading, it looks like going through bounded-variation function methods will be easier, but I'm still curious if this is a good approach (since, all else equal, it's still true). – Eugene Shvarts Jan 17 '13 at 20:53
up vote 1 down vote accepted

Let us call $$D^+ \nu (x) := \limsup_{\varepsilon \to 0} {\frac{\nu(]x-\varepsilon,x+\varepsilon[)}{2\varepsilon}},\, D^- \nu (x) := \liminf_{\varepsilon \to 0} {\frac{\nu(]x-\varepsilon,x+\varepsilon[)}{2\varepsilon}}.$$ You can show with the help of the Vitali covering theorem, that for a measurable set $A \subseteq [0,1]$ we have $$D^+ \nu(x) \geq q \quad \forall\, x \in A \qquad \Longrightarrow \qquad \nu(A) \geq q \lambda(A).$$ Now consider $A_n := \{x \in K \: | \: D^+ \nu(x) \geq 1/n\}$. Then we get $\nu(A_n) \geq \lambda(A_n)/n$, but the left-hand side is $0$, so $\lambda(A_n) = 0$ and by taking the limit $\lambda(\{x \in K \: | \: D^+ \nu(x) > 0\}) = 0$, which means $\lambda(\{x \in K \: | \: D \nu(x) = 0\}) = \lambda(K) - 0 = 1$. So the derivative exists almost everywhere and is zero where it exists.

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Thanks Thomas! It's interesting to see things done the measure-theoretic way. – Eugene Shvarts Jan 17 '13 at 22:43

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