6 views

Skip to first unread message

Nov 24, 2006, 9:23:31 AM11/24/06

to

Let f:[a,b]-> R. Characterize f that satisfy m(f(E))=0 for every set E with

m(E)=0,

where m is the Lebesgue measure.

For example, every absolutely continuous

monotonic function satisfies this condition, but certainly the class of

functions

satisfying this condition is much larger.

m(E)=0,

where m is the Lebesgue measure.

For example, every absolutely continuous

monotonic function satisfies this condition, but certainly the class of

functions

satisfying this condition is much larger.

Nov 24, 2006, 10:09:05 AM11/24/06

to

"TCL" <tl...@verizon.net> wrote in message

news:DfD9h.15826$yV.5558@trnddc07...

continuous.

Nov 24, 2006, 11:32:05 AM11/24/06

to

This is not correct. A function need not even be continuous to have

this property. For example, a step function will have this property.

Also, any function whose range has measure 0 will certainly have this

property, and this class of function contains non-measurable functions.

Nov 24, 2006, 11:36:50 AM11/24/06

to

"Jules" <julia...@gmail.com> wrote in message

news:1164385925....@f16g2000cwb.googlegroups.com...

is the characterization? If this is too hard, let us say f is continuous,

then what is the characterization?

Nov 24, 2006, 8:39:12 PM11/24/06

to

TCL wrote:

This is usually called the "Lusin (N)" condition. Note that

"Lusin" is sometimes spelled "Luzin". It's usually applied

to measurable functions.

http://www.google.com/search?q=Lusin-N+function

http://www.google.com/search?q=Luzin-N+function

James Foran done some research on N-functions.

Although I'm sure others have done more, this suggests

looking at his text "Fundamentals of Real Analysis",

and when I did, I see some results on pp. 351-???.

What follows are a few results from some measure theory

notes I wrote up 20 years ago:

Any function Lipschitz on an interval is an N-function.

[This can be proved directly by the characterization of

measure zero in terms of coverings by open intervals.]

A continuous function is an N-function if and only if

it maps measurable sets to measurable sets.

[For (==>), use the fact that any measurable set is a

union of an F_sigma set and a set of measure zero set,

and observe that the continuous image of an F_sigma set

is an F_sigma set (because any F_sigma set is a countable

union of compact sets). For (<==), work with a non-measurable

subset of a positive measure image of a measure zero set.]

If f and g are N-functions, then 'f compose g' is an N-function,

but not necessarily f + g.

If f is absolutely continuous on [a,b], then f is an N-function.

[This can be proved using the characterization of absolute

continuity that makes use of countable (and not just finite)

collections of non-overlapping intervals. This is Exercise #14

in Chapter 5 of Royden, 2'nd edition.]

If P is a non-measurable set, then f is a non-measurable

N-function, where f(x) = x + 1 if x in P, and f(x) = x if

x not in P.

If f is an N-function, continuous on [a,b], and has bounded

variation on [a,b], then f is absolutely continuous on [a,b].

[Hewitt/Stromberg, 18.25, pp. 288-290; Natanson (Vol. I),

pp. 250-252.]

We can't omit BV[a,b] in the previous result. Let C be the

Cantor middle thirds set and let {(a_n, b_n): n = 1, 2, ...}

be its bounded complementary open intervals, enumerated in

some fixed order. Let c_n = (1/2)(a_n + b_n). Define

f:[0,1] --> R by: f(x) = 0 if x in C; f(x) = 1/n if x = c_n;

f(x) is linear in each [a_n, c_n] and [c_n, b_n].

There exists a continuous N-function f:[0,1] --> R such that

f' doesn't exist on a set of positive measure.

[Similar to above, except use a fat Cantor set in [0,1] for C,

replace 1/n with (1/2)(b_n - a_n) + d_n, where d_n is the maximum

length of the intervals that are a subset of [0,1] intersect

the complement of UNION(k=1 to n) of [a_k, b_k]. See 18.40

on p. 297 of Hewitt/Stromberg.]

Dave L. Renfro

Nov 25, 2006, 10:10:45 AM11/25/06

to

Dave L. Renfro wrote (in part):

> If f is an N-function, continuous on [a,b], and has bounded

> variation on [a,b], then f is absolutely continuous on [a,b].

> [Hewitt/Stromberg, 18.25, pp. 288-290; Natanson (Vol. I),

> pp. 250-252.]

Apparently, this is called the Banach-Zarecki theorem.

I was looking something up in Bruckner/Bruckner/Thomson's

1997 text "Real Analysis" and saw this as Theorem 7.11

(p. 293). The statement in BBT says that f is absolutely

continuous on [a,b] IF AND ONLY IF, on [a,b], f is continuous,

is BV, and is an N-function.

Also, googling "Banach-Zarecki" shows that this label for the

result is used elsewhere, although I have no idea how standard

the term is.

Some useful results about N-functions can also be found on

pp. 154-160 (and perhaps some later pages as well) of

Rooij/Schikhof's 1982 book "A Second Course on Real Functions".

Dave L. Renfro

Nov 26, 2006, 9:37:11 AM11/26/06

to

"Dave L. Renfro" <renf...@cmich.edu> wrote in message

news:1164418752....@f16g2000cwb.googlegroups.com...

> If f and g are N-functions, then 'f compose g' is an N-function,

> but not necessarily f + g.

Do you have examples to show that f+g need not be an N-function?

Nov 26, 2006, 12:06:38 PM11/26/06

to

TCL wrote:

> Do you have examples to show that f+g need not be an N-function?

An example is outlined in Exercise 21G (p. 155) of Rooij/Schikhof's

"A Second Course in Real Functions". Incidentally, Saks' "Theory

of the Integral", 2'nd edition, has a discussion of Lusin's

Condition (N) in Chapter VII, Section 6 (pp. 224-228). Saks

mentions a paper by N. Bary [1] as a place to find an extensive

discussion of these functions. I glanced over it (table of

contents on p. 188, N-functions introduced on pp. 194-201),

along with another paper of Bary's [2], for an f + g example,

but I didn't see one. However, I can't really read French,

so if you can, you should look at them yourself.

[1] N. Bary, Mathematische Annalen 103 (1930), 185-248 & 598-653

http://dz-srv1.sub.uni-goettingen.de/cache/toc/D37021.html

[2] N. Bary, C. R. Acad. Sci. Paris 189 (1929), 441-443.

ftp://ftp.bnf.fr/000/N0003142_PDF_441_443.pdf

I thought the journal "Real Analysis Exchange" had published

a survey article on N-functions, but I glanced through the

table of contents of the past volumes (which I have, and

survey articles are easy to spot because they are published

under the heading of "Survey Articles" and always at the top

of the table of contents) and didn't see one. I'm sure there

are several papers in this journal that deal with N-functions,

but I didn't take the time to look for them.

Anyway, here's Rooij/Schikhof's example.

Let C be the Cantor middle thirds set in [0,1]. If f:C --> R

is continuous such that f[C] has measure zero, then f can be

extended to a function on [0,1] that has property (N).

[R/S's hint: Make it differentiable outside of C.]

We can define a sequence of continuous functions g_1, g_2, ...

from C to the set {0, 1} by

x = 2 * SUM(k=1 to oo) of [ (g_k)(x) * 3^(-k) ]

for all x in C.

There exist continuous functions f_1 and f_2 from [0,1]

to R, each with property (N), such that

(f_1)(x) = SUM(k=1 to oo) of [ (g_{2k})(x) * 3^(-k) ]

(f_2)(x) = SUM(k=1 to oo) of [ (g_{2k+1})(x) * 3^(-k) ]

for all x in C.

One can show that f_1 + f_2 does not have property (N)

by making use of the fact that

C + C =(def.) {c + c': c,c' in C} = [0,2]

(i.e. the sum-set of the Cantor set is the interval [0,2]).

Dave L. Renfro

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu