D**u
发帖数: 204 | 1 I am wondering whether there exists such a "measure" F on ALL the subsets of the set of
integers, which satisfies the following conditions:
(1) F(S) = 0 for any finite subset S.
(2) F(S) can be negative.
(3) finite additivity: If S is the union of 2 disjoint subsets S1 and S2,
then F(S) = F(S1) + F(S2).
(4) non-triviality: F(S) != 0 for some S. | t*****n
发帖数: 225 | 2 Define a finitely additive measure F on all the subsets
of the set Z of integers as follows:
For any subset S of integers, any positive integer n,
define f_S(n)=1/n if n is in S, and f_S(n)=0 if n is not in S.
Then f_S1+f_S2= f_S if S is the union of 2 disjoint subsets S1 and S2.
Define F(S)=lim (f_S(1)+...+f_S(n))/log n,
then
1. F(S)=0 for any finite subset S.
2. F(S) is non-negative.
3. F(S) is finitely additive.
4. F(Z)=1.
楼主发包子吧
of the set of
【在 D**u 的大作中提到】
: I am wondering whether there exists such a "measure" F on ALL the subsets of the set of
: integers, which satisfies the following conditions:
: (1) F(S) = 0 for any finite subset S.
: (2) F(S) can be negative.
: (3) finite additivity: If S is the union of 2 disjoint subsets S1 and S2,
: then F(S) = F(S1) + F(S2).
: (4) non-triviality: F(S) != 0 for some S.
| Q***5
发帖数: 994 | 3 Interesting construction. But you need to prove that F is well defined for
every S -- this does not seem to be true because the limit does not exist
for some set.
【在 t*****n 的大作中提到】
: Define a finitely additive measure F on all the subsets
: of the set Z of integers as follows:
: For any subset S of integers, any positive integer n,
: define f_S(n)=1/n if n is in S, and f_S(n)=0 if n is not in S.
: Then f_S1+f_S2= f_S if S is the union of 2 disjoint subsets S1 and S2.
: Define F(S)=lim (f_S(1)+...+f_S(n))/log n,
: then
: 1. F(S)=0 for any finite subset S.
: 2. F(S) is non-negative.
: 3. F(S) is finitely additive.
| t*****n
发帖数: 225 | 4 you can replace limit by a Banach limit or
lim_{\omega} used in the definition of Dixmier trace.
See wikipedia for the definition of Banach limit or Dixmier trace.
unfortunately, according to wikipedia, there seems no explicit
construction of a Banach limit, although it exists if you accept
the axiom of choice.
Interesting construction. But you need to prove that F is well defined for
every S -- this does not seem to be true because the limit does not exist
for some set.
【在 Q***5 的大作中提到】
: Interesting construction. But you need to prove that F is well defined for
: every S -- this does not seem to be true because the limit does not exist
: for some set.
| Q***5
发帖数: 994 | 5 Checked wikipedia, I think you are right, a non trivial Banach limit should
do the job.
I was thinking about construct such a measure directly, but if axiom of
choice has to be involved -- I give up -- thanks for saving my time.
【在 t*****n 的大作中提到】
: you can replace limit by a Banach limit or
: lim_{\omega} used in the definition of Dixmier trace.
: See wikipedia for the definition of Banach limit or Dixmier trace.
: unfortunately, according to wikipedia, there seems no explicit
: construction of a Banach limit, although it exists if you accept
: the axiom of choice.
:
: Interesting construction. But you need to prove that F is well defined for
: every S -- this does not seem to be true because the limit does not exist
: for some set.
| D**u
发帖数: 204 | 6 Very nice construction, thanks.
Noticed that "log(N)" appears in the Dixmier trace wiki page, but not in the
Banach limit wiki page, so I am wondering if the following construction
also works?
f_S(n)=1 if n is in S, otherwise 0.
F(S) is the Banach limit of sequence ((f_S(1)+...+f_S(n))/n).
【在 t*****n 的大作中提到】
: you can replace limit by a Banach limit or
: lim_{\omega} used in the definition of Dixmier trace.
: See wikipedia for the definition of Banach limit or Dixmier trace.
: unfortunately, according to wikipedia, there seems no explicit
: construction of a Banach limit, although it exists if you accept
: the axiom of choice.
:
: Interesting construction. But you need to prove that F is well defined for
: every S -- this does not seem to be true because the limit does not exist
: for some set.
| j******w
发帖数: 690 | 7 Here is an example:
For any S\subseteq \Omega, F(S)=0 if and only if S is not cofinite. Otherwis
e, F(S)=1.
of the set of
【在 D**u 的大作中提到】
: I am wondering whether there exists such a "measure" F on ALL the subsets of the set of
: integers, which satisfies the following conditions:
: (1) F(S) = 0 for any finite subset S.
: (2) F(S) can be negative.
: (3) finite additivity: If S is the union of 2 disjoint subsets S1 and S2,
: then F(S) = F(S1) + F(S2).
: (4) non-triviality: F(S) != 0 for some S.
| D**u
发帖数: 204 | 8 There is a problem here.
F((2*n)) + F((2*n+1)) = F(Z), so F((2*n)) and F((2*n+1)) can not be 0 the
same time.
Otherwis
【在 j******w 的大作中提到】
: Here is an example:
: For any S\subseteq \Omega, F(S)=0 if and only if S is not cofinite. Otherwis
: e, F(S)=1.
:
: of the set of
| t*****n
发帖数: 225 | 9 Yes, it works as well. I was using log n only to imitate Dixmier trace which
has exactly the properties as your question requires.
After that I also realized the construction can be made simpler.
the
for
exist
【在 D**u 的大作中提到】
: Very nice construction, thanks.
: Noticed that "log(N)" appears in the Dixmier trace wiki page, but not in the
: Banach limit wiki page, so I am wondering if the following construction
: also works?
: f_S(n)=1 if n is in S, otherwise 0.
: F(S) is the Banach limit of sequence ((f_S(1)+...+f_S(n))/n).
|
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