Functions that take sets of measure zero to sets of measure zero

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TCL

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


TCL

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Nov 24, 2006, 10:09:05 AM11/24/06
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"TCL" <tl...@verizon.net> wrote in message
news:DfD9h.15826$yV.5558@trnddc07...
I think the answer is: f has this property if and only if f is absolutely
continuous.


Jules

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

TCL

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Nov 24, 2006, 11:36:50 AM11/24/06
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"Jules" <julia...@gmail.com> wrote in message
news:1164385925....@f16g2000cwb.googlegroups.com...
Sure, you are right. Let us say we require that f be measurable. Then what
is the characterization? If this is too hard, let us say f is continuous,
then what is the characterization?


Dave L. Renfro

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

Dave L. Renfro

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

TCL

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


Dave L. Renfro

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

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