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Nov 6, 2006, 1:38:43 PM11/6/06

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Hello

Suppose f:I -> R is differentiable on an open interval I of R. Let D be

the set of points where f is discontinuous. We know D is meager and,

therefore, has an empty interior. I have 2 questions, about which I

couldn't come to a conclusion:

a) Is it possible that D is dense in I? Since f has the intermediate

value property, it seems to me it can't, but I couldn't come to a

conclusion.

b) D can have positive Lebesgue measure, right?

Thank you.

Amanda

Nov 6, 2006, 2:23:39 PM11/6/06

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Amanda wrote:

I think you meant "... where f' is discontinuous" in the

first paragraph and "Since f' has the ..." in the second

paragraph.

The continuity set of a derivative on an open interval J

is dense in J. In fact, the continuity set has cardinality c

in every subinterval of J. On the other hand, the discontinuity

set of a derivative can have the following properties --->

1. dense in the reals

2. cardinality c in every interval

3. positive measure (hence, not Riemann integrable)

4. positive measure in every interval

5. full measure in every interval (i.e. measure zero complement)

6. have a Hausdorff dimension zero complement

Briefly, a subset D of the reals is a discontinuity set for some

bounded derivative if and only if D is an F_sigma first category

set. Moreover, derivatives for which D is large are plentiful in

the sense of Baire category (use the sup norm on the collection

of bounded derivatives) -- For most bounded derivatives, the set D

is the complement of a measure zero set. Note that this is much

stronger than simply saying that D has positive measure (i.e. that

the derivative isn't Riemann integrable). Note also that D and

the complement of D, for any of these Baire-typical bounded

derivatives, gives a partition of the reals into a measure zero set

and a first category set. In 1993 Bernd Kirchheim strengthened

this by proving that, given any Hausdorff measure function h,

for most bounded derivatives the set D is the complement of

a set that has Hausdorff h-measure zero. (I seem to have

left out Kirchheim's result in the first post below.)

HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES

http://groups.google.com/group/sci.math/msg/814be41b1ea8c024

Proof that derivatives have the intermediate value property

http://mathforum.org/kb/thread.jspa?messageID=5259756

A couple of years ago I was hard at work on a revision of

the "historical essay" post above, but then I had to stop and

work on other things (such as my Fall classes, which began

before I had finished). I haven't gotten back to it since

then, but it's something I intend on returning to eventually.

The revision would be several times longer and more detailed,

and what I've done so far is over twice as long already.

Dave L. Renfro

Nov 6, 2006, 3:13:53 PM11/6/06

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In article

<1162838322.9...@f16g2000cwb.googlegroups.com>,

"Amanda" <sc...@hotmail.com> wrote:

<1162838322.9...@f16g2000cwb.googlegroups.com>,

"Amanda" <sc...@hotmail.com> wrote:

> Hello

>

> Suppose f:I -> R is differentiable on an open interval I of R. Let D be

> the set of points where f is discontinuous.

You mean f' in that last line.

> We know D is meager and,

> therefore, has an empty interior. I have 2 questions, about which I

> couldn't come to a conclusion:

>

> a) Is it possible that D is dense in I?

Yes. A basic theorem here is: Suppose f_n is a sequence of

differentiable functions on [a,b]. If f_n, f_n' converge

uniformly to f, g resp. on [a,b], then f' = g on [a,b].

So given a countable set D = {d_1, ...} in R, we can choose

functions f_n, differentiable everywhere, such that f_n' is

continuous everywhere except at d_n, and such that |f_n|, |f_n'|

< 1/2^n on R. f = f_1 + f_2 + ... will be differentiable on R,

and f' will be discontinuous precisely on D. D of course could be

dense in R.

> Since f has the intermediate

> value property, it seems to me it can't, but I couldn't come to a

> conclusion.

>

> b) D can have positive Lebesgue measure, right?

Yes, it can have full measure. Meaning, we can have R = D U E,

where m(E) = 0. The idea would be to take D = U K_n, where the

K_n's are pairwise disjoint compact sets with no interior (aka

fat Cantor sets). If you set f_n(x) =

d(x,K_n)^2*sin(1/d(x,K_n))*2^(-n), and add them up, you'll get an

example. Almost. Maybe not quite. You'd have to play around with

it a little.

> Thank you.

> Amanda

Nov 8, 2006, 5:11:07 PM11/8/06

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Thank you all for your help

Amanda

Amanda

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