Mixing
(T/F) If $mu$ is a Borel measure on $mathbb{R}$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all...
1
$begingroup$
I am trying to prove or disprove the following statement:
If $mu$ is a Borel measure on $mathbb{R}$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.
I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.
measure-theory borel-sets
asked Jan 28 at 5:44
johnny133253johnny133253
470111
$endgroup$
add a comment |
1
$begingroup$
I am trying to prove or disprove the following statement:
If $mu$ is a Borel measure on $mathbb{R}$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.
I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.
measure-theory borel-sets
asked Jan 28 at 5:44
johnny133253johnny133253
470111
$endgroup$
$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11
add a comment |
1
1
1
0
$begingroup$
I am trying to prove or disprove the following statement:
If $mu$ is a Borel measure on $mathbb{R}$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.
I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.
measure-theory borel-sets
asked Jan 28 at 5:44
johnny133253johnny133253
470111
$endgroup$
I am trying to prove or disprove the following statement:
If $mu$ is a Borel measure on $mathbb{R}$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.
I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.
measure-theory borel-sets
measure-theory borel-sets
asked Jan 28 at 5:44
johnny133253johnny133253
470111
asked Jan 28 at 5:44
johnny133253johnny133253
470111
asked Jan 28 at 5:44
johnny133253johnny133253
470111
asked Jan 28 at 5:44
johnny133253johnny133253
470111
asked Jan 28 at 5:44
johnny133253johnny133253
470111
470111
$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11
add a comment |
$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11
$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
$$
mu(A)=lim_{ntoinfty}mu(Acap[-n,n])=0.
$$
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
$endgroup$
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
add a comment |
1
$begingroup$
Note by countable additivity of measure,
$0 leq mu(A)= sum_{nin mathbb Z} mu(A cap (n,n+1]) leq sum_{nin mathbb Z} mu(A cap [n,n+1])= 0$
Hence $mu(A)=0$.
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
$endgroup$
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
$$
mu(A)=lim_{ntoinfty}mu(Acap[-n,n])=0.
$$
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
$endgroup$
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
add a comment |
2
$begingroup$
$$
mu(A)=lim_{ntoinfty}mu(Acap[-n,n])=0.
$$
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
$endgroup$
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
add a comment |
2
2
2
$begingroup$
$$
mu(A)=lim_{ntoinfty}mu(Acap[-n,n])=0.
$$
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
$endgroup$
$$
mu(A)=lim_{ntoinfty}mu(Acap[-n,n])=0.
$$
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
edited Jan 28 at 6:10
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
answered Jan 28 at 6:02
d.k.o.d.k.o.
10.4k630
10.4k630
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
add a comment |
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]cup[2,3]$ compact, for instance?
$endgroup$
– johnny133253
Jan 28 at 6:09
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
$begingroup$
math.stackexchange.com/questions/234292/…
$endgroup$
– d.k.o.
Jan 28 at 6:11
1
1
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
You stated that $mu(A cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$.
$endgroup$
– d.k.o.
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
$begingroup$
Got it. Thank you.
$endgroup$
– johnny133253
Jan 28 at 6:14
add a comment |
1
$begingroup$
Note by countable additivity of measure,
$0 leq mu(A)= sum_{nin mathbb Z} mu(A cap (n,n+1]) leq sum_{nin mathbb Z} mu(A cap [n,n+1])= 0$
Hence $mu(A)=0$.
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
$endgroup$
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
add a comment |
1
$begingroup$
Note by countable additivity of measure,
$0 leq mu(A)= sum_{nin mathbb Z} mu(A cap (n,n+1]) leq sum_{nin mathbb Z} mu(A cap [n,n+1])= 0$
Hence $mu(A)=0$.
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
$endgroup$
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
add a comment |
1
1
1
$begingroup$
Note by countable additivity of measure,
$0 leq mu(A)= sum_{nin mathbb Z} mu(A cap (n,n+1]) leq sum_{nin mathbb Z} mu(A cap [n,n+1])= 0$
Hence $mu(A)=0$.
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
$endgroup$
Note by countable additivity of measure,
$0 leq mu(A)= sum_{nin mathbb Z} mu(A cap (n,n+1]) leq sum_{nin mathbb Z} mu(A cap [n,n+1])= 0$
Hence $mu(A)=0$.
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
edited Jan 28 at 7:17
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
answered Jan 28 at 6:19
Mayuresh LMayuresh L
1,331323
1,331323
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
add a comment |
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
Missing $mu$? Also the second "$=$" should be "$le$"...
$endgroup$
– d.k.o.
Jan 28 at 6:27
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
$begingroup$
I missed it. Thanks.
$endgroup$
– Mayuresh L
Jan 28 at 7:15
add a comment |
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$begingroup$
Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:52
$begingroup$
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra.
$endgroup$
– johnny133253
Jan 28 at 5:54
$begingroup$
Thank you for the reply.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 5:58
$begingroup$
Every $sigma$-finite Borel measure on a Polish space is inner regular for compacta.
$endgroup$
– Henno Brandsma
Jan 28 at 10:11