Non zero Lebesgue measure
Diogo Aguiar Gomes
graph, i.e. the set S={(x,f(x)): x in R} doesn't have zero Lebesgue
Measure. It seems to me that the set S is non mensurable (if exists) but
I don't know if such a function really exists.
Thanks
Diogo Gomes
Robert Israel
|> In article <4cgmc6$b...@ci.ist.utl.pt>, Diogo Aguiar Gomes
|> The inner measure is strictly less than the outer measure.
|> Let f be the characteristic function of this set. Of course
|> the graph is non-measurable. But it has positive 2-dimensional
|> Lebesgue outer measure.
??? This graph is contained in the union of the two lines y=1 and y=0, so
it has 2-dimensional Lebesgue measure 0.
Consider the following example:
Let S be the family of all sets of finite 2-dimensional measure of the form
union_{i=1}^infinity (a_i, b_i) x (c_i, d_i)
where a_i, b_i, c_i and d_i are rational.
Then S has the cardinality of the continuum.
For each member U of S, the set V(U) of x such that the vertical line through
(x,0) is not a subset of U has cardinality of the continuum (in fact contains
all reals except a set of measure 0).
Now define a 1-1 function g: S -> R such that g(U) is in V(U) for all U in S
(to do this, you can take a 1-1 correspondence between S and the least ordinal
of cardinality of the continuum, and use transfinite induction).
Finally, take f: R -> R so that if x = g(U), (x, f(x)) is not in U (define
f(x) arbitrarily if x is not in g(S)).
Let G be the graph of f. If it had zero 2-dimensional Lebesgue measure,
there would be some U in S that contained it. But by construction,
(g(U), f(g(U))) is not in U.
--
Robert Israel isr...@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Y4
Philo D.
> Does anyone know an example of a function f:R->R, such that its
> graph, i.e. the set S={(x,f(x)): x in R} doesn't have zero Lebesgue
> Measure. It seems to me that the set S is non mensurable (if exists) but
> I don't know if such a function really exists.
Consider a non-measurable subset of the interval [0,1].
The inner measure is strictly less than the outer measure.
Let f be the characteristic function of this set. Of course
the graph is non-measurable. But it has positive 2-dimensional
Lebesgue outer measure.
--
*
That was obviously another Antilipe. [A Walk Through H]
Ilias Kastanas
Robert Israel <isr...@math.ubc.ca> wrote:
>In article <doozy-04019...@mw520ma.mps.ohio-state.edu>, do...@dizzy.mps.ohio-state.edu (Philo D.) writes:
>|> <dgo...@math.ist.utl.pt> wrote:
>|>
>|> > Does anyone know an example of a function f:R->R, such that its
>|> > graph, i.e. the set S={(x,f(x)): x in R} doesn't have zero Lebesgue
>|> > Measure. It seems to me that the set S is non mensurable (if exists) but
>|> > I don't know if such a function really exists.
>|>
Kinkier than thou
-----------------
The graph of a measurable function is measurable... not only for
Lebesgue measure on the real line and plane, but even for a general measu-
rable space as domain (and the reals carrying a Borel measure as range).
There is even a related approach to integration... "area under the curve",
taken seriously!
It is not difficult to make a function's graph "big". Consider so-
lutions of f(x+y) = f(x) + f(y). Continuous (even measurable ones) are
f(x) = cx. But there are discontinuous ones: Let H be a (Hamel) basis,
for the reals as a vector space over the rationals. So every x is (unique-
ly!) a finite sum q_1*h_1 + .. + q_k*h_k, q's rational. Then f(x) =
q_1 + ... +q_k is as desired. Its graph is dense on the plane.
Lebesgue gave a function on [0,1], whose range on any subinterval
is [0,1]! Let x = 0.a_1 a_2 a_3... (decimal expansion). If its "odd half"
0.a_1 a_3 a_5... is irrational, f(x) = 0. If it is rational with the
first repeating part at a_2n-1, f(x) = 0.a_2n a_2n+2 a_2n+4...
Its graph is dense, of course.
After warming up, we now define a function on R, yes, (-inf, inf),
so that its range on any subinterval is (-inf, inf)... and yet it is
also zero almost everywhere!
Start with the Cantor function ( 1/2 on the middle third of [0,1],
1/4 and 3/4 on the next left and right middle thirds, etc; a continuous,
monotonic function (with derivative = 0 almost everywhere), c(x).
Now g(x) = tan( pi(c(x) - 1/2)), 0 < x < 1, maps the Cantor set C
( intersected with (0, 1) ) onto R, i.e. (-inf, inf). We can squeeze it
into any interval (a, b): a subset Z_ab, of measure 0, defined by a +
+ (b - a)*x, x in C, has range R under g_ab(x) = g(x-a / b-a), x in Z_ab.
Next we arm ourselves with lots of copies of Z/g... Define f(x) = 0
on the integers. Put a copy of Z_ab in every (n, n+1), and define f to be
g_ab accordingly. The set V_1 where f is still undefined is open, hence
a union of disjoint intervals... put a Z_ab with g_ab in each. Continue
this with V_2, ... That's countably many batches of Z_ab 's, each of
measure 0. Let f be 0 everywhere else. Given any interval, some interval
of some V_k manages to burrow in. And f(x) is 0 almost everywhere -- hence
measurable.
Variations on such pathology are now easy to come by. Of course, for
measurable f the graph cannot avoid having plane Lebesgue measure 0, due
to Fubini's theorem. All vertical sections are singletons!
A non-Lebesgue-measurable graph needs the Axiom of Choice for its
construction, so we don't hold back. A variation of the Bernstein set
construction produces a non-measurable set all of whose vertical and
horizontal sections are at most singletons. Cosider [0,1] X [0,1]. There
are c (continuum) many compact sets of positive measure... so wellorder
them! Note that any closed uncountable set has cardinality c. Pick some
(x_0, y_0). Having chosen points for ordinals up to a, consider the a+1-th
set K...By Fubini, it has a vertical section of positive measure at a
"fresh" x, =/= x_i for i up to a. For the same reason one of the (x, y) in
K has a fresh y. Let (x, y) be the a+1 choice.
The resulting set S is not measurable; if it were, by Fubini it would
have measure 0, and its complement measure 1; but its complement fails to
contain any closed (= compact) set of positive measure.
Crass use of choice, sure; but it is inevitable, so it's OK.
Ilias