Talk:Singular function

message: Singular function mentions it as a (strictly increasing) singular function; However, according to Minkowski's question mark function, it is differentiable only on the rationals, not almost everywhere, as is required by the definition here. Am I missing something? -- Meni Rosenfeld (talk) 00:16, 28 October 2007 (UTC)
replies: text: Minkowski's question mark function says "The derivative vanishes on the rational numbers" which is similar to the position here.  It also says "It does not have a well-defined derivative, in the classical sense, on the irrationals; however, there are several constructions for a measure that, when integrated, yields the question mark function" which is not the same as here, and makes Minkowski's question mark function slippery. --Rumping (talk) 17:06, 8 January 2008 (UTC)
thread_title:  Minkowski's question mark function 

message: Heh. Regarding the above conversation, I believe that careful analysis shows that the question mark has non-zero derivative on a set of non-zero measure. However, I've seen literature refer to it as a "singular function" even though it clearly doesn't meet the definition given in this article.  Which means that either there's a common usage that is incorrect, or that there is some broader definition of singular function that allows for a non-zero-derivative on a set of measure greater than zero. Also, FWIW, I have never seen any sort of proof that the circle-map winding number has non-zero derivative only on a set of measure zero -- in fact, its probably *not* a set of measure zero: as K is varied from 0 to 1, the flat parts of circle map winding number get wider i.e. the measure of the flat parts gets bigger, and, at K=0, the measure of the flat bits is exactly zero (since there's no phase-locking at all).
Last but not least: only the "canonical" Cantor set has a measure of zero, but there are other Cantor sets one can construct with measure that is greater than zero (e.g. instead of removing the middle 1/3 each time, one removes geometrically less each time. This results in a fat Cantor set, for which one can define a distribution function which looks like the devils staircase, but fails to meet the strict definition of this article. 
Thus, we have three different examples that seem to be "devil's staircases", but don't seem to meet the "measure zero" part of the current definition given in this article. I suspect the "measure zero" part of the definition given here is wrong. Seems to me the measure-zero requirement is not germane to the definition -- rather, that the proper definition probably should be "the derivative is non-zero only on a cantor set (not necessarily the canonical cantor set), and that the measure of the set where the derivative is non-vanishing could be anything, from zero to one, thus allowing both the question mark, and erasing doubts about whether the circle-map winding number is or is not "singular".  Alternately, perhaps the correct definition of a singular function is *any* function that behaves like a measure, and is a singular measure with respect to the Lebesgue measure.  linas (talk) 05:17, 11 February 2010 (UTC)
thread_title: Measure Zero!?!?

message: While it's ok to interpret singular functions as cdf, 
the arguments given for why it's neither a discrete nor
a continuous distribution sound wrong to me:
"...neither a discrete random variable (since the probability density is zero everywhere it exists)..." 
It's not a discrete random variable because for any single point x, its
probability is zero, P(X=x)=0. This is a consequence of f being continuous
(and therefore left-continuous). It's not a consequence of the probability density
(presumably pdf) being zero; in fact, discrete random variables all have
the property that the derivative of their cdf is zero wherever it exists
in the classical sense.
"...nor an absolutely continuous random variable (since the probability is zero for each point)..."
A continuous random variable would have zero probability for each individual point, P(X=x)=0,
so this property does not contradict the random variable being continuous -- it is actually
a property of continuous random variables. Rather,
the fact that the pdf is zero almost everywhere shows that the random variable defined
by the cdf f (where f is singular), is not a continuous random variable, i.e.
because the associated pdf integrates up to zero.
A rephrasing might be
" neither a discrete random variable (since the probability is zero for each point) ..."
"...nor an absolutely continuous random variable (since the probability density is zero
almost everywhere)..."
Comments?  —Preceding unsigned comment added by 128.40.24.185 (talk) 15:03, 5 October 2010 (UTC)
thread_title:  Probabilistic Interpretation