Outer measure
Mate
I,
where m is the usual Lebesgue outer measure.
Is it true that w = m ?
G. A. Edgar
<1e11bbb4-709f-45fc...@g10g2000yqh.googlegroups.com>,
Mate <mma...@personal.ro> wrote:
On the m-measurable sets, the two outer measures agree. But for some
non-m-measurable sets, w many be smaller than m.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Chip Eastham
wrote:
> In article
> <1e11bbb4-709f-45fc-9e34-7d9d96f76...@g10g2000yqh.googlegroups.com>,
>
> Mate <mmat...@personal.ro> wrote:
> > Let w be an outer measure in R such that w(I) = m(I)
> for each interval
> > I,
> > where m is the usual Lebesgue outer measure.
> > Is it true that w = m ?
>
> On the m-measurable sets, the two outer measures agree.
> But for some
> non-m-measurable sets, w many be smaller than m.
Hi, Dr. Edgar:
I'm a little unclear what you mean that "w [may]
be smaller than m" for some non-m-measurable sets
(m is the usual Lebesgue measure in R and w is an
outer measure agreeing with m on intervals).
If set S is not m-measurable, we can't compare the
values w(S) and m(S). If on the other hand you
are comparing the domains of definition for w and
m, to say w is "smaller" than m would seem to
mean all w-measurable sets are m-measurable.
tia, chip
Gc
The OP said that the m is a outer measure.
Chip Eastham
Thanks, that clarifies things a little for me.
You are correct that OP said "m is the usual Lebesgue
outer measure". However it seems to me that this would
mean a (countably subadditive) measure defined for all
subsets of R by m(S) = inf(SUM(m(I)) where the sums are
taken over all finite covers of S by intervals I, and
the infimum is taken over all such covers.
So "w may be smaller than m" could be true if taken in
the sense of domains of definition, because m is defined
for all subsets of R, and then I'm confused only about
distinguishing cases of "m-measurable sets" and "non-
m-measurable sets".
regards, chip
Mate
> > Let w be an outer measure in R such that w(I) = m(I) for each interval
> > I,
> > where m is the usual Lebesgue outer measure.
> > Is it true that w = m ?
>
> On the m-measurable sets, the two outer measures agree. But for some
> non-m-measurable sets, w many be smaller than m.
>
> --
> G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Thank you for the answer, which is NO as I understand.
Can you suggest a method to obtain such a w ?
Mate
G. A. Edgar
<94a44575-5162-4891...@r1g2000yqj.googlegroups.com>,
Chip Eastham <hard...@gmail.com> wrote:
> On Mar 12, 10:30�am, "G. A. Edgar" <ed...@math.ohio-state.edu.invalid>
> wrote:
> > In article
> > <1e11bbb4-709f-45fc-9e34-7d9d96f76...@g10g2000yqh.googlegroups.com>,
> >
> > Mate <mmat...@personal.ro> wrote:
> > > Let w be an outer measure in R such that w(I) = m(I)
> > for each interval
> > > I,
> > > where m is the usual Lebesgue outer measure.
> > > Is it true that w = m ?
> >
> > On the m-measurable sets, the two outer measures agree.
> > �But for some
> > non-m-measurable sets, w many be smaller than m.
>
> Hi, Dr. Edgar:
>
> I'm a little unclear what you mean that "w [may]
> be smaller than m" for some non-m-measurable sets
> (m is the usual Lebesgue measure in R and w is an
> outer measure agreeing with m on intervals).
>
> If set S is not m-measurable, we can't compare the
> values w(S) and m(S).
You said "outer measure" so I took it that they are defined
for ALL sets. And I did mean that for non-measurable set A
(where Lebesgue inner measure is < Lebesgue outer measure),
it is possible that w(A) is some value strictly between
the Lebesgue inner and outer measures of A.
> If on the other hand you
> are comparing the domains of definition for w and
> m, to say w is "smaller" than m would seem to
> mean all w-measurable sets are m-measurable.
>
> tia, chip
--
G. A. Edgar
<2f2fb766-c481-449f...@g10g2000yqh.googlegroups.com>,
Mate <mma...@personal.ro> wrote:
Let A \subset [0,1] have inner measure 0 and outer measure 1.
(When I don't specify a measure, I mean Lebesgue.)
We can define a measure w on the sigma-algebra generated by the
measurable sets plus A that agrees with m on the measurable
sets and has w(A) = 1/2, say. Use this w to define an outer measure.
Then w and m agree on all m-measurable sets, but w(A) = 1/2 < 1 = m(A).
Gc
countable covers.
> the infimum is taken over all such covers.
Yes.
> So "w may be smaller than m" could be true if taken in
> the sense of domains of definition, because m is defined
No, because w was also an outer measure. If u is an outer measure on
R^n,
we say that some set A is u-measurable, if it satisfies the
Caratheodory condition.
u(E) = u(A /\ E) + u(E\A), for every subset E \in R^n. Then the
restriction of u to those sets A etc. is a "proper" measure.
Mate
wrote:
> In article
> <2f2fb766-c481-449f-abce-2fd4d1025...@g10g2000yqh.googlegroups.com>,
Thank you again. I found the details of this construction in Halmos'
Measure Theory, p. 71.
Mate
David C. Ullrich
<ed...@math.ohio-state.edu.invalid> wrote:
So I conjectured yesterday, but I didn't see exactly how to define w.
Duh. It's easier to prove something that you know is true...
For the benefit of anyone who's wondering about this:
Say B is the complement of A. To save typing, the letter "E"
below will refer to a Lebesgue measurable set and the letter
"S" to a set in the larger sigma-algebra.
The first "duh" is that the larger algebra consists precisely of
sets of the form S = E, S = E union A and S = E union B.
(because it's easy to see that the sets of this form form
a sigma-algebra).
Now (assuming we're talking about measures on all of R)
the second duh is that a trivial definition of w works:
Define w(E) = E, w(E union A) = m(E) + 1/2 and
w(E union B) = infinity. I didn't see how that could
work since E and A need not be disjoint. But the
countable additivity of w is trivial:
Say S = union S_j where the S_j are disjoint.
If one of the S_i is E_j union B then S is E union B
and w(S) = infinity, sum w(S_j) = infinity + something = infinity.
If none of the S_j are E_j union B then we note (duh!)
that at most one of them is E_j union A. If we have
S_j = E_j for all j fine, and if there exists one j with
S_j = E_j union A then S = (union E_j) union A and
so w(S) = m(union E_j) + 1/2 = 1/2 + sum m(E_j)
= sum w(E_j).
I thought it should be harder than this. Duh.
Chip Eastham
wrote:
> In article
> <94a44575-5162-4891-847f-a1a727121...@r1g2000yqj.googlegroups.com>,
Thanks, and to Gc above as well.
--c
G. A. Edgar
Ullrich <ull...@math.okstate.edu> wrote:
> The first "duh" is that the larger algebra consists precisely of
> sets of the form S = E, S = E union A and S = E union B.
> (because it's easy to see that the sets of this form form
> a sigma-algebra).
Perhaps not closed under intersections... E cap A for example ...
but in fact there is a simple description:
(E cap A) union (F cap B) where E and F are Lebesgue measurable.
David C. Ullrich
<ed...@math.ohio-state.edu.invalid> wrote:
>In article <ue4np55k2b393km4q...@4ax.com>, David C.
>Ullrich <ull...@math.okstate.edu> wrote:
>
>> The first "duh" is that the larger algebra consists precisely of
>> sets of the form S = E, S = E union A and S = E union B.
>> (because it's easy to see that the sets of this form form
>> a sigma-algebra).
>
>Perhaps not closed under intersections... E cap A for example ...
>but in fact there is a simple description:
>
>(E cap A) union (F cap B) where E and F are Lebesgue measurable.
Eewps, back to the drawing board...
David C. Ullrich
<ed...@math.ohio-state.edu.invalid> wrote:
>In article <ue4np55k2b393km4q...@4ax.com>, David C.
>Ullrich <ull...@math.okstate.edu> wrote:
>
>> The first "duh" is that the larger algebra consists precisely of
>> sets of the form S = E, S = E union A and S = E union B.
>> (because it's easy to see that the sets of this form form
>> a sigma-algebra).
>
>Perhaps not closed under intersections... E cap A for example ...
>but in fact there is a simple description:
>
>(E cap A) union (F cap B) where E and F are Lebesgue measurable.
Took me a minute to see what the definition of w should be...
Let's restrict everything to [0,1]; what happens in the complement
doesn't matter. In particular, B is [0,1] \ A.
Then B also has inner measure 0 and outer measure 1.
So we define
w((E cap A) union (F cap B) = (1/2) (m(E) + m(F)).
Seems wacky but I believe it works. First, w is well-defined:
Say (E cap A) union (F cap B) = (E' cap A) union (F' cap B).
Then E cap A = E' cap A, so the symmetric difference of E and
E' is a measurable subset of B, which must have measure 0;
hence m(E) = m(E'). Similarly m(F) = m(F').
It's easy to verify that w(A) = 1/2 and w(E) = m(E).
And countable additivity follows from a trivial lemma:
Lemma. If E_n cap E_m has measure zero for n <> m
then m(union E_n) = sum m(E_n).
Now say (E_n cap A) union (F_n cap B) are pairwise
disjoint. Then E_n cap E_m is a measurable subset
of B for n <> m, hence has measure zero. So the
lemma shows that the measure of the union of the E_n
is the sum of the measures; similarly for the F_n
and hence w(union ((E_n cap A) union (F_n cap B))
is the sum of the w's.