An example $limsup mu_n(F)<mu(F)$
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Give examples of a sequence of measures $(mu)_ninmathbb{N}$ which converges on weak topology for $mu$, on a measurable topological space $Omega,mathscr{F}$ and a closed set $Finmathscr{F}$ which $limsup mu_n(F)<mu(F)$
I do not know if the following example is right:$delta_{-frac{1}{n}}([−1,0])$
$lim_{ntoinfty}delta_{-frac{1}{n}}([−1,0])=delta_0$
For $n>Ninmathbb{N}$ $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$
which implies $limsup delta_{-frac{1}{n}}([−1,0])<delta_{0}([−1,0])=0$
Question:
Is this right?
If not what could be an alternative?
Thanks in advance!
measure-theory proof-writing
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
$endgroup$
add a comment |
$begingroup$
Give examples of a sequence of measures $(mu)_ninmathbb{N}$ which converges on weak topology for $mu$, on a measurable topological space $Omega,mathscr{F}$ and a closed set $Finmathscr{F}$ which $limsup mu_n(F)<mu(F)$
I do not know if the following example is right:$delta_{-frac{1}{n}}([−1,0])$
$lim_{ntoinfty}delta_{-frac{1}{n}}([−1,0])=delta_0$
For $n>Ninmathbb{N}$ $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$
which implies $limsup delta_{-frac{1}{n}}([−1,0])<delta_{0}([−1,0])=0$
Question:
Is this right?
If not what could be an alternative?
Thanks in advance!
measure-theory proof-writing
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
$endgroup$
$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05
add a comment |
$begingroup$
Give examples of a sequence of measures $(mu)_ninmathbb{N}$ which converges on weak topology for $mu$, on a measurable topological space $Omega,mathscr{F}$ and a closed set $Finmathscr{F}$ which $limsup mu_n(F)<mu(F)$
I do not know if the following example is right:$delta_{-frac{1}{n}}([−1,0])$
$lim_{ntoinfty}delta_{-frac{1}{n}}([−1,0])=delta_0$
For $n>Ninmathbb{N}$ $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$
which implies $limsup delta_{-frac{1}{n}}([−1,0])<delta_{0}([−1,0])=0$
Question:
Is this right?
If not what could be an alternative?
Thanks in advance!
measure-theory proof-writing
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
$endgroup$
Give examples of a sequence of measures $(mu)_ninmathbb{N}$ which converges on weak topology for $mu$, on a measurable topological space $Omega,mathscr{F}$ and a closed set $Finmathscr{F}$ which $limsup mu_n(F)<mu(F)$
I do not know if the following example is right:$delta_{-frac{1}{n}}([−1,0])$
$lim_{ntoinfty}delta_{-frac{1}{n}}([−1,0])=delta_0$
For $n>Ninmathbb{N}$ $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$
which implies $limsup delta_{-frac{1}{n}}([−1,0])<delta_{0}([−1,0])=0$
Question:
Is this right?
If not what could be an alternative?
Thanks in advance!
measure-theory proof-writing
measure-theory proof-writing
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
asked Jan 17 at 10:59
Pedro GomesPedro Gomes
1,8222721
1,8222721
$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05
add a comment |
$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05
$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05
$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05
add a comment |
1 Answer
1
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If I understand your question correctly, you can take the sequence $mu_n:=delta_{1/n}$ which converges weak$^*$-ly to $mu=delta_0$ (in the topology of $C_0(Bbb R)^*$). Let $F:={0}$, then
$$
limsup_{ntoinfty} delta_{1/n}(F) = 0 < 1 = delta_0(F).
$$
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I understand your question correctly, you can take the sequence $mu_n:=delta_{1/n}$ which converges weak$^*$-ly to $mu=delta_0$ (in the topology of $C_0(Bbb R)^*$). Let $F:={0}$, then
$$
limsup_{ntoinfty} delta_{1/n}(F) = 0 < 1 = delta_0(F).
$$
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
If I understand your question correctly, you can take the sequence $mu_n:=delta_{1/n}$ which converges weak$^*$-ly to $mu=delta_0$ (in the topology of $C_0(Bbb R)^*$). Let $F:={0}$, then
$$
limsup_{ntoinfty} delta_{1/n}(F) = 0 < 1 = delta_0(F).
$$
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
If I understand your question correctly, you can take the sequence $mu_n:=delta_{1/n}$ which converges weak$^*$-ly to $mu=delta_0$ (in the topology of $C_0(Bbb R)^*$). Let $F:={0}$, then
$$
limsup_{ntoinfty} delta_{1/n}(F) = 0 < 1 = delta_0(F).
$$
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
$endgroup$
If I understand your question correctly, you can take the sequence $mu_n:=delta_{1/n}$ which converges weak$^*$-ly to $mu=delta_0$ (in the topology of $C_0(Bbb R)^*$). Let $F:={0}$, then
$$
limsup_{ntoinfty} delta_{1/n}(F) = 0 < 1 = delta_0(F).
$$
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:09
BigbearZzzBigbearZzz
8,69121652
8,69121652
add a comment |
add a comment |
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$begingroup$
Why is $delta_0[-1,0]>delta_{-frac{1}{n}}([−1,0])$? Aren't they both $1$?
$endgroup$
– BigbearZzz
Jan 17 at 11:05