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cardinality of Lebesque measurable and Borel sets

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lite.o...@gmail.com

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Jan 20, 2009, 9:13:41 PM1/20/09
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How to prove that cardinality of lebesgue measurable sets (in R) is
2^c?

How to prove cardinality of borel sets (in R) is c?

Stephen J. Herschkorn

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Jan 20, 2009, 10:24:11 PM1/20/09
to lite.o...@gmail.com, Stephen J. Herschkorn
lite.o...@gmail.com wrote:

Is this homework? What have you tried so far?

--
Stephen J. Herschkorn sjher...@netscape.net
Math Tutor on the Internet and in Central New Jersey

W^3

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Jan 20, 2009, 11:40:59 PM1/20/09
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In article
<ac5e71ec-664b-43bc...@w39g2000prb.googlegroups.com>,
lite.o...@gmail.com wrote:

> How to prove that cardinality of lebesgue measurable sets (in R) is
> 2^c?

Hint: You've undoubtedly met a "large" set of measure 0.

lite.o...@gmail.com

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Jan 21, 2009, 12:59:23 AM1/21/09
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On Jan 20, 10:24 pm, "Stephen J. Herschkorn" <sjhersc...@netscape.net>
wrote:

> lite.on.b...@gmail.com wrote:
> >How to prove that cardinality of lebesgue measurable sets (in R) is
> >2^c?
>
> >How to prove cardinality of borel sets (in R) is c?
>
> Is this homework?  What have you tried so far?
>

Yes homework.

Have tried things that have no hope of working (like setting up
injections from Reals).

Fatal

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Jan 21, 2009, 3:00:45 AM1/21/09
to
lite.o...@gmail.com a écrit :

> How to prove cardinality of borel sets (in R) is c?

You can look at the notion of Suslin subsets of R.

--
Fatal

G. A. Edgar

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Jan 21, 2009, 6:19:44 AM1/21/09
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In article <4976d62e$0$15129$426a...@news.free.fr>, Fatal
<fa...@yahoo.fr> wrote:

> lite.o...@gmail.com a écrit :
>
> > How to prove cardinality of borel sets (in R) is c?
>

More generally, a countably generated sigma-algebra can have cardinal
at most c

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

David C. Ullrich

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Jan 21, 2009, 6:28:37 AM1/21/09
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On Tue, 20 Jan 2009 18:13:41 -0800 (PST), lite.o...@gmail.com
wrote:

>
>How to prove that cardinality of lebesgue measurable sets (in R) is
>2^c?

It's clear that the cardinality is <= 2^c, right? And ___ has
cardinality c and measure 0, hence every subset of ___ is
Lebesgue measurable...

>How to prove cardinality of borel sets (in R) is c?

The only way I can think of uses what I think of as the
"bottom-up" construction of the sigma-algebra generated
by a set S of subsets of R (as opposed to "consider the
intersection of all the sigma-algebras containing S",
which is the "top-down" construction). I don't
know how much set theory you know so I don't
know whether the following will make sense:

Say S is a subset of the power set of S. Assume that
the empty set and R are in S for convenience.

Let S_0 = S. For each countable ordinal alpha,
let S_alpha be the set of all complements and
countable unions of the union of S_beta for
beta < alphs. If w_1 is the first uncountable
ordinal then S_w_1 is the sigma-algebra
generated by S.

Now is S is the set of open subsets of R you show
that S has cardinality c and then you show by
induction on alphs that S_alpha has cardinality
c for every alpha...

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)

José Carlos Santos

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Jan 21, 2009, 7:33:00 AM1/21/09
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On 21-01-2009 11:28, David C. Ullrich wrote:

>> How to prove cardinality of borel sets (in R) is c?
>
> The only way I can think of uses what I think of as the
> "bottom-up" construction of the sigma-algebra generated
> by a set S of subsets of R (as opposed to "consider the
> intersection of all the sigma-algebras containing S",
> which is the "top-down" construction). I don't
> know how much set theory you know so I don't
> know whether the following will make sense:
>
> Say S is a subset of the power set of S.

Of course, the second S should be R. Otherwise, this would be one of the
worst choices of notation in existence.

Best regards,

Jose Carlos Santos

Dave L. Renfro

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Jan 21, 2009, 10:12:55 AM1/21/09
to
Fatal wrote:

>> How to prove cardinality of borel sets (in R) is c?

G. A. Edgar wrote:

> More generally, a countably generated sigma-algebra
> can have cardinal at most c

Unfortunately (especially for me, as you'll see) there
seems to be a problem with this argument. I don't know
where/if you saw this argument, but if it was in my sci.math
post below (see [1]) and/or in David Bressoud's recent real
analysis book (see pp. 290 & 321 of [2]), where he cites
my sci.math post for this approach, then there is likely
a problem with this argument.

[1] sci.math -- Borel set (21 May 2007)
http://groups.google.com/group/sci.math/msg/66168cf580929605

[2] David M. Bressoud, "A Radical Approach to Lebesgue's Theory
of Integration", Mathematical Association of America, 2008.

On 11 March 2008, Klaas Pieter Hart sent me an e-mail saying
essentially: (I'm going to paraphrase a bit, because I haven't
asked his permission to post his exact comments.)

He doesn't see how the argument works. Take a countable
generating collection for the Borel sets. This generating
collection has c many subcollections. However, he doesn't
see an obvious injection from the collection of Borel sets
to the power set of this generating collection.

Later that same day I sent him the following reply:

-----------------------

It certainly seems that I'm mistaken. At best, it's not
obvious (now) how the fact I cite can be used to prove
there are at most c many Borel sets. Given a Borel set E,
if I denote by C_E a countable subcollection of a fixed
basis of the Borel sets such that E belongs to sigma(C_E),
then I see no reason why we should have the property
"E not equal to F implies that sigma(C_E) not equal to
sigma(C_F)". Indeed, it seems worse than this, since I
see no reason why C_E is not equal to C_F.

I'll think about this some more, but I'm not optimistic
that I'm overlooking something now, as opposed to having
overlooked something when I wrote that post.

-----------------------

Dave L. Renfro

Stephen J. Herschkorn

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Jan 21, 2009, 12:39:42 PM1/21/09
to lite.o...@gmail.com, Stephen J. Herschkorn
lite.o...@gmail.com wrote:

The former is easier: Every set of measure 0 is a Lebesgue set.

For the latter, build up the Borel sets by transfinite induction.

David C. Ullrich

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Jan 22, 2009, 6:28:44 AM1/22/09
to
On Wed, 21 Jan 2009 07:12:55 -0800 (PST), "Dave L. Renfro"
<renf...@cmich.edu> wrote:

>Fatal wrote:
>
>>> How to prove cardinality of borel sets (in R) is c?
>
>G. A. Edgar wrote:
>
>> More generally, a countably generated sigma-algebra
>> can have cardinal at most c
>
>Unfortunately (especially for me, as you'll see) there
>seems to be a problem with this argument. I don't know
>where/if you saw this argument, but if it was in my sci.math
>post below (see [1]) and/or in David Bressoud's recent real
>analysis book (see pp. 290 & 321 of [2]), where he cites
>my sci.math post for this approach, then there is likely
>a problem with this argument.
>
>[1] sci.math -- Borel set (21 May 2007)
> http://groups.google.com/group/sci.math/msg/66168cf580929605

Which argument are you referring to, and why do you assume
it's the argument Edgar had in mind? There are two arguments
sketched in that post...

Below it seems you're referring to the second argument(?).
Are you also claiming there's an problem with the argument
by transfinite induction?

David C. Ullrich

Dave L. Renfro

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Jan 22, 2009, 10:26:21 AM1/22/09
to
David C. Ullrich wrote (in part):

> Which argument are you referring to, and why do you assume
> it's the argument Edgar had in mind? There are two arguments
> sketched in that post...
>
> Below it seems you're referring to the second argument(?).
> Are you also claiming there's an problem with the argument
> by transfinite induction?

There's nothing wrong with the argument by transfinite
induction over the Borel hierarchy -- it's in all the
texts (well, those that get into this -- mostly the old classical
real analysis and topology texts and the newer undergraduate
level set theory texts). However, back around 2003 I thought
I had come up with a different way to show there are at
most c Borel sets (it's trivial there are at least c Borel
sets; indeed, any sigma-algebra is either finite or has
cardinality at least c), and used the opportunity of my 2005
sci.math post to mention it. I was sure the other "proof" was
fairly well known, but then I began to notice that no other
proof (besides the one making use of the transfinite Borel
hierarchy) seemed to be in the literature. In particular,
I didn't come across it any any of Borel's, Baire's, Hausdorff's,
Lebesgue's, Sierpinski's, and other mathematicians' papers
and books. This surprised me, especially given all the work
on sigma-algebra notions done in the 1920s and 1930s
(for example, all the Russians' work on C-sets and R-sets).

Anyway, at the moment I know of no other proof that there
are at least c Borel sets than the proof that makes explicit
use of the Borel hierarchy.

Dave L. Renfro

Dave L. Renfro

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Jan 22, 2009, 10:46:40 AM1/22/09
to
Dave L. Renfro wrote (in part):

> cardinality at least c), and used the opportunity of my 2005
> sci.math post to mention it. I was sure the other "proof" was

Ooops! This should have been "my 2007 sci.math post". The other
date I gave, 2003 for when I came up with the idea and wrote it
down in my notes, is correct (if anyone cares, which I doubt).

Dave L. Renfro

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