Support of a Lebesgue-Stieltjes measure [closed] The Next CEO of Stack OverflowProbability measure with predefined supportmeasure of the boundary of the supportRelation between support of image-measure and closure of the imageTwo measures are equivalent iff their image measures have same support?Lebesgue-Stieltjes measureWhy is any Lebesgue-Stieltjes measure on $(mathbbR,mathcalB_mathbbR)$ not complete?Stieltjes measure function to Lebesgue Measure?Question about the Support of a Discrete Probability Measuremeasure of “support of measure”2-dimensional Lebesgue-Stieltjes measure (Bivariate Density to Distribution Function)
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Support of a Lebesgue-Stieltjes measure [closed]
The Next CEO of Stack OverflowProbability measure with predefined supportmeasure of the boundary of the supportRelation between support of image-measure and closure of the imageTwo measures are equivalent iff their image measures have same support?Lebesgue-Stieltjes measureWhy is any Lebesgue-Stieltjes measure on $(mathbbR,mathcalB_mathbbR)$ not complete?Stieltjes measure function to Lebesgue Measure?Question about the Support of a Discrete Probability Measuremeasure of “support of measure”2-dimensional Lebesgue-Stieltjes measure (Bivariate Density to Distribution Function)
$begingroup$
Let $G:[0,1]to[0,1]$ be a distribution function (that is, non-decreasing and right-continuous) with support $textsupp(G)$ define as:
$$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;forall varepsilon>0.$$
Associated with $G$, there exists a Lebesgue-Stieltjes probability measure $mu_G$ s.t. for $I=(a,b]$ with $0<a<ble 1$, $mu_G(I)=G(b)-G(a)$. The support of $mu_G$ is the smallest closed set with measure 1.
- Show that $textsupp(mu_G)=textsupp(G)$.
- If $Sin[0,1]backslash textsupp(G)$, then $mu_G(S)=0$, but the converse not necessarily true. Under what conditions does the converse hold?
measure-theory probability-distributions lebesgue-measure
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
$endgroup$
closed as off-topic by K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer Mar 20 at 6:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer
add a comment |
$begingroup$
Let $G:[0,1]to[0,1]$ be a distribution function (that is, non-decreasing and right-continuous) with support $textsupp(G)$ define as:
$$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;forall varepsilon>0.$$
Associated with $G$, there exists a Lebesgue-Stieltjes probability measure $mu_G$ s.t. for $I=(a,b]$ with $0<a<ble 1$, $mu_G(I)=G(b)-G(a)$. The support of $mu_G$ is the smallest closed set with measure 1.
- Show that $textsupp(mu_G)=textsupp(G)$.
- If $Sin[0,1]backslash textsupp(G)$, then $mu_G(S)=0$, but the converse not necessarily true. Under what conditions does the converse hold?
measure-theory probability-distributions lebesgue-measure
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
$endgroup$
closed as off-topic by K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer Mar 20 at 6:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer
$begingroup$
What is your definition of $operatornamesupp(G)$?
$endgroup$
– user251257
Mar 19 at 22:32
$begingroup$
$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
$endgroup$
– Caio Lorecchio
Mar 19 at 22:37
$begingroup$
your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
$endgroup$
– user251257
Mar 19 at 23:07
$begingroup$
Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 19 at 23:24
$begingroup$
your reasoning looks okay. Anyway I posted an answer with a (different) proof.
$endgroup$
– user251257
Mar 19 at 23:49
add a comment |
$begingroup$
Let $G:[0,1]to[0,1]$ be a distribution function (that is, non-decreasing and right-continuous) with support $textsupp(G)$ define as:
$$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;forall varepsilon>0.$$
Associated with $G$, there exists a Lebesgue-Stieltjes probability measure $mu_G$ s.t. for $I=(a,b]$ with $0<a<ble 1$, $mu_G(I)=G(b)-G(a)$. The support of $mu_G$ is the smallest closed set with measure 1.
- Show that $textsupp(mu_G)=textsupp(G)$.
- If $Sin[0,1]backslash textsupp(G)$, then $mu_G(S)=0$, but the converse not necessarily true. Under what conditions does the converse hold?
measure-theory probability-distributions lebesgue-measure
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
$endgroup$
Let $G:[0,1]to[0,1]$ be a distribution function (that is, non-decreasing and right-continuous) with support $textsupp(G)$ define as:
$$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;forall varepsilon>0.$$
Associated with $G$, there exists a Lebesgue-Stieltjes probability measure $mu_G$ s.t. for $I=(a,b]$ with $0<a<ble 1$, $mu_G(I)=G(b)-G(a)$. The support of $mu_G$ is the smallest closed set with measure 1.
- Show that $textsupp(mu_G)=textsupp(G)$.
- If $Sin[0,1]backslash textsupp(G)$, then $mu_G(S)=0$, but the converse not necessarily true. Under what conditions does the converse hold?
measure-theory probability-distributions lebesgue-measure
measure-theory probability-distributions lebesgue-measure
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
edited Mar 20 at 11:53
Caio Lorecchio
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
asked Mar 19 at 21:24
Caio LorecchioCaio Lorecchio
365
365
closed as off-topic by K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer Mar 20 at 6:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer
closed as off-topic by K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer Mar 20 at 6:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – K.Power, Shailesh, Alex Provost, Leucippus, Eevee Trainer
$begingroup$
What is your definition of $operatornamesupp(G)$?
$endgroup$
– user251257
Mar 19 at 22:32
$begingroup$
$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
$endgroup$
– Caio Lorecchio
Mar 19 at 22:37
$begingroup$
your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
$endgroup$
– user251257
Mar 19 at 23:07
$begingroup$
Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 19 at 23:24
$begingroup$
your reasoning looks okay. Anyway I posted an answer with a (different) proof.
$endgroup$
– user251257
Mar 19 at 23:49
add a comment |
$begingroup$
What is your definition of $operatornamesupp(G)$?
$endgroup$
– user251257
Mar 19 at 22:32
$begingroup$
$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
$endgroup$
– Caio Lorecchio
Mar 19 at 22:37
$begingroup$
your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
$endgroup$
– user251257
Mar 19 at 23:07
$begingroup$
Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 19 at 23:24
$begingroup$
your reasoning looks okay. Anyway I posted an answer with a (different) proof.
$endgroup$
– user251257
Mar 19 at 23:49
$begingroup$
What is your definition of $operatornamesupp(G)$?
$endgroup$
– user251257
Mar 19 at 22:32
$begingroup$
What is your definition of $operatornamesupp(G)$?
$endgroup$
– user251257
Mar 19 at 22:32
$begingroup$
$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
$endgroup$
– Caio Lorecchio
Mar 19 at 22:37
$begingroup$
$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
$endgroup$
– Caio Lorecchio
Mar 19 at 22:37
$begingroup$
your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
$endgroup$
– user251257
Mar 19 at 23:07
$begingroup$
your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
$endgroup$
– user251257
Mar 19 at 23:07
$begingroup$
Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 19 at 23:24
$begingroup$
Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 19 at 23:24
$begingroup$
your reasoning looks okay. Anyway I posted an answer with a (different) proof.
$endgroup$
– user251257
Mar 19 at 23:49
$begingroup$
your reasoning looks okay. Anyway I posted an answer with a (different) proof.
$endgroup$
– user251257
Mar 19 at 23:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For Item 1:
For brevity, let $S_1 = operatornamesupp(mu_G)$, $S_2 = operatornamesupp(G)$, and $S_3 = rin[0,1] : forall Utext open in [0, 1] text with rin U implies mu_G(U) > 0 $.
I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $mu_G(S_1) = 1$. Let $rin S_1$ and $U$ open with $rin U$. Then, $S_1setminus U$ is closed and thus we have $mu_G(S_1 setminus U) < 1$. In particular, we have
$$mu_G(U) ge mu_G(S_1 cap U) = 1 - mu_G(S_1 setminus U) > 0.$$
On other hand, let $rin S_3$ and $C$ be closed with $mu_G(C) = 1$. Assume $rnotin C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies
$$ mu_G(C cup U) = mu_G(C) + mu_G(U) > 1,$$
which contradicts that $mu_G([0, 1]) = 1$.Proof for $S_2subseteq S_3$:
Let $rin S_2$ and $varepsilon > 0$. Then, we have
$$ 0 < mu_G((r - varepsilon, r + varepsilon)) le mu_G((r - varepsilon, r + varepsilon]) = G(r+varepsilon) - G(r - varepsilon) $$
and therefore $rin S_3$.Proof for $S_3subseteq S_2$:
Now, let $rin S_3$, $U$ be open with $rin U$, and $varepsilon > 0$ with $(r-varepsilon, r+varepsilon) subseteq U$. Then, we have
$$ 0 < G(r+varepsilon/2) - G(r - varepsilon/2) = mu_G((r - varepsilon / 2, r + varepsilon / 2]) le mu_G((r - varepsilon, r + varepsilon)) le mu_G(U) $$
and therefore $rin S_2$.
For Item 2:
The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.
answered Mar 19 at 23:02
user251257user251257
7,64721128
$endgroup$
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
$endgroup$
– Caio Lorecchio
Mar 20 at 0:40
$begingroup$
@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
$endgroup$
– user251257
Mar 20 at 0:46
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 20 at 2:56
$begingroup$
@CaioLorecchio yes. $$
$endgroup$
– user251257
Mar 20 at 3:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For Item 1:
For brevity, let $S_1 = operatornamesupp(mu_G)$, $S_2 = operatornamesupp(G)$, and $S_3 = rin[0,1] : forall Utext open in [0, 1] text with rin U implies mu_G(U) > 0 $.
I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $mu_G(S_1) = 1$. Let $rin S_1$ and $U$ open with $rin U$. Then, $S_1setminus U$ is closed and thus we have $mu_G(S_1 setminus U) < 1$. In particular, we have
$$mu_G(U) ge mu_G(S_1 cap U) = 1 - mu_G(S_1 setminus U) > 0.$$
On other hand, let $rin S_3$ and $C$ be closed with $mu_G(C) = 1$. Assume $rnotin C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies
$$ mu_G(C cup U) = mu_G(C) + mu_G(U) > 1,$$
which contradicts that $mu_G([0, 1]) = 1$.Proof for $S_2subseteq S_3$:
Let $rin S_2$ and $varepsilon > 0$. Then, we have
$$ 0 < mu_G((r - varepsilon, r + varepsilon)) le mu_G((r - varepsilon, r + varepsilon]) = G(r+varepsilon) - G(r - varepsilon) $$
and therefore $rin S_3$.Proof for $S_3subseteq S_2$:
Now, let $rin S_3$, $U$ be open with $rin U$, and $varepsilon > 0$ with $(r-varepsilon, r+varepsilon) subseteq U$. Then, we have
$$ 0 < G(r+varepsilon/2) - G(r - varepsilon/2) = mu_G((r - varepsilon / 2, r + varepsilon / 2]) le mu_G((r - varepsilon, r + varepsilon)) le mu_G(U) $$
and therefore $rin S_2$.
For Item 2:
The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.
answered Mar 19 at 23:02
user251257user251257
7,64721128
$endgroup$
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
$endgroup$
– Caio Lorecchio
Mar 20 at 0:40
$begingroup$
@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
$endgroup$
– user251257
Mar 20 at 0:46
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 20 at 2:56
$begingroup$
@CaioLorecchio yes. $$
$endgroup$
– user251257
Mar 20 at 3:17
add a comment |
$begingroup$
For Item 1:
For brevity, let $S_1 = operatornamesupp(mu_G)$, $S_2 = operatornamesupp(G)$, and $S_3 = rin[0,1] : forall Utext open in [0, 1] text with rin U implies mu_G(U) > 0 $.
I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $mu_G(S_1) = 1$. Let $rin S_1$ and $U$ open with $rin U$. Then, $S_1setminus U$ is closed and thus we have $mu_G(S_1 setminus U) < 1$. In particular, we have
$$mu_G(U) ge mu_G(S_1 cap U) = 1 - mu_G(S_1 setminus U) > 0.$$
On other hand, let $rin S_3$ and $C$ be closed with $mu_G(C) = 1$. Assume $rnotin C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies
$$ mu_G(C cup U) = mu_G(C) + mu_G(U) > 1,$$
which contradicts that $mu_G([0, 1]) = 1$.Proof for $S_2subseteq S_3$:
Let $rin S_2$ and $varepsilon > 0$. Then, we have
$$ 0 < mu_G((r - varepsilon, r + varepsilon)) le mu_G((r - varepsilon, r + varepsilon]) = G(r+varepsilon) - G(r - varepsilon) $$
and therefore $rin S_3$.Proof for $S_3subseteq S_2$:
Now, let $rin S_3$, $U$ be open with $rin U$, and $varepsilon > 0$ with $(r-varepsilon, r+varepsilon) subseteq U$. Then, we have
$$ 0 < G(r+varepsilon/2) - G(r - varepsilon/2) = mu_G((r - varepsilon / 2, r + varepsilon / 2]) le mu_G((r - varepsilon, r + varepsilon)) le mu_G(U) $$
and therefore $rin S_2$.
For Item 2:
The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.
answered Mar 19 at 23:02
user251257user251257
7,64721128
$endgroup$
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
$endgroup$
– Caio Lorecchio
Mar 20 at 0:40
$begingroup$
@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
$endgroup$
– user251257
Mar 20 at 0:46
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
$endgroup$
– Caio Lorecchio
Mar 20 at 2:56
$begingroup$
@CaioLorecchio yes. $$
$endgroup$
– user251257
Mar 20 at 3:17
add a comment |
$begingroup$
For Item 1:
For brevity, let $S_1 = operatornamesupp(mu_G)$, $S_2 = operatornamesupp(G)$, and $S_3 = rin[0,1] : forall Utext open in [0, 1] text with rin U implies mu_G(U) > 0 $.
I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $mu_G(S_1) = 1$. Let $rin S_1$ and $U$ open with $rin U$. Then, $S_1setminus U$ is closed and thus we have $mu_G(S_1 setminus U) < 1$. In particular, we have
$$mu_G(U) ge mu_G(S_1 cap U) = 1 - mu_G(S_1 setminus U) > 0.$$
On other hand, let $rin S_3$ and $C$ be closed with $mu_G(C) = 1$. Assume $rnotin C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies
$$ mu_G(C cup U) = mu_G(C) + mu_G(U) > 1,$$
which contradicts that $mu_G([0, 1]) = 1$.Proof for $S_2subseteq S_3$:
Let $rin S_2$ and $varepsilon > 0$. Then, we have
$$ 0 < mu_G((r - varepsilon, r + varepsilon)) le mu_G((r - varepsilon, r + varepsilon]) = G(r+varepsilon) - G(r - varepsilon) $$
and therefore $rin S_3$.Proof for $S_3subseteq S_2$:
Now, let $rin S_3$, $U$ be open with $rin U$, and $varepsilon > 0$ with $(r-varepsilon, r+varepsilon) subseteq U$. Then, we have
$$ 0 < G(r+varepsilon/2) - G(r - varepsilon/2) = mu_G((r - varepsilon / 2, r + varepsilon / 2]) le mu_G((r - varepsilon, r + varepsilon)) le mu_G(U) $$
and therefore $rin S_2$.
For Item 2:
The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.
answered Mar 19 at 23:02
user251257user251257
7,64721128
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For Item 1:
For brevity, let $S_1 = operatornamesupp(mu_G)$, $S_2 = operatornamesupp(G)$, and $S_3 = rin[0,1] : forall Utext open in [0, 1] text with rin U implies mu_G(U) > 0 $.
I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $mu_G(S_1) = 1$. Let $rin S_1$ and $U$ open with $rin U$. Then, $S_1setminus U$ is closed and thus we have $mu_G(S_1 setminus U) < 1$. In particular, we have
$$mu_G(U) ge mu_G(S_1 cap U) = 1 - mu_G(S_1 setminus U) > 0.$$
On other hand, let $rin S_3$ and $C$ be closed with $mu_G(C) = 1$. Assume $rnotin C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies
$$ mu_G(C cup U) = mu_G(C) + mu_G(U) > 1,$$
which contradicts that $mu_G([0, 1]) = 1$.Proof for $S_2subseteq S_3$:
Let $rin S_2$ and $varepsilon > 0$. Then, we have
$$ 0 < mu_G((r - varepsilon, r + varepsilon)) le mu_G((r - varepsilon, r + varepsilon]) = G(r+varepsilon) - G(r - varepsilon) $$
and therefore $rin S_3$.Proof for $S_3subseteq S_2$:
Now, let $rin S_3$, $U$ be open with $rin U$, and $varepsilon > 0$ with $(r-varepsilon, r+varepsilon) subseteq U$. Then, we have
$$ 0 < G(r+varepsilon/2) - G(r - varepsilon/2) = mu_G((r - varepsilon / 2, r + varepsilon / 2]) le mu_G((r - varepsilon, r + varepsilon)) le mu_G(U) $$
and therefore $rin S_2$.
For Item 2:
The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.
answered Mar 19 at 23:02
user251257user251257
7,64721128
edited Mar 20 at 0:44
answered Mar 19 at 23:02
user251257user251257
7,64721128
answered Mar 19 at 23:02
user251257user251257
7,64721128
answered Mar 19 at 23:02
user251257user251257
7,64721128
7,64721128
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
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– Caio Lorecchio
Mar 20 at 0:40
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@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
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– user251257
Mar 20 at 0:46
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I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
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– Caio Lorecchio
Mar 20 at 2:56
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@CaioLorecchio yes. $$
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– user251257
Mar 20 at 3:17
add a comment |
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Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
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– Caio Lorecchio
Mar 20 at 0:40
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@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
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– user251257
Mar 20 at 0:46
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
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– Caio Lorecchio
Mar 20 at 2:56
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@CaioLorecchio yes. $$
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– user251257
Mar 20 at 3:17
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
$endgroup$
– Caio Lorecchio
Mar 20 at 0:40
$begingroup$
Thanks! You define $textsupp(mu_G)$ simply as the smallest closed subset $S$ of $[0,1]$ s.t. $mu_G(S)=1$, right? With respect to item 1, I was looking for some conditions to say that a set in $textsupp(G)$ has certainly a positive measure, or maybe some conditions to say that a set has positive measure if and only if it belongs to the support. I was thinking about the Lebesgue measure, where each singleton has zero measure, but it belongs to the support (the real line).
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– Caio Lorecchio
Mar 20 at 0:40
$begingroup$
@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
$endgroup$
– user251257
Mar 20 at 0:46
$begingroup$
@CaioLorecchio: Oh my bad. I got confused with the $S_i$'s. I fixed them. For Item 2, as you said, one need to impose some conditions on $S$.
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– user251257
Mar 20 at 0:46
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
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– Caio Lorecchio
Mar 20 at 2:56
$begingroup$
I got the point anyway, thanks :). I think you changed the subscripts in steps 2 and 3. In step 2, you want to show that $S_3subseteq S_2$ and in step 3, you want to show that $S_2subseteq S_3$, right? One last question. It is sufficient that $S$ has non-empty interior because if that is the case, $mu_G(S)ge mu_G(textint(S))>0$, since $textint(S)$ is open and any point in it belongs to the support. Is that correct?
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– Caio Lorecchio
Mar 20 at 2:56
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@CaioLorecchio yes. $$
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– user251257
Mar 20 at 3:17
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@CaioLorecchio yes. $$
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– user251257
Mar 20 at 3:17
add a comment |
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What is your definition of $operatornamesupp(G)$?
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– user251257
Mar 19 at 22:32
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$textsupp(G)=rin [0,1]: G(r+varepsilon)-G(r-varepsilon)>0;;forall varepsilon>0$ and $textsupp(mu_G)=capCin mathcalC: mu_G(C)=1$, where $mathcalC$ is the collection of closed subsets of $[0,1]$. I have managed to show that $textsupp(G)$ is a closed subset, so it belongs to $mathcalC$.
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– Caio Lorecchio
Mar 19 at 22:37
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your definition of $operatornamesupp(mu_G)$ makes the thing difficult. Do you know the equivalence to $operatornamesupp(mu_G) = r : mu_G(U) > 0 ; forall U text open with rin U $?
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– user251257
Mar 19 at 23:07
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Yes. If $textsupp(mu_G)$ is the intersection of all closed subsets of $[0,1]$ with measure 1, then $textsupp(G)^c=cupUin mathcalU: mu_G(U)=0$ where $mathcalU$ is the collection of open sets of $[0,1]$. But this equals $r: mu_G(U)=0;textfor some U open with $rin U$$. Setting $(textsupp(mu_G)^c)^c=textsupp(mu_G)$ gives the result. Is that correct?
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– Caio Lorecchio
Mar 19 at 23:24
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your reasoning looks okay. Anyway I posted an answer with a (different) proof.
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– user251257
Mar 19 at 23:49