A measure zero uncountable set for the Lebesgue-Stieltjes measure $mu_F$.
$begingroup$
Let $mu_F$ be the Lebesgue-Stieltjes measure on $Bbb R$ associated with the increasing function $F:Bbb Rto Bbb R$. Construct an uncountable set of measure 0 for $mu_F$.
When $F(x)=x$ and we get the Lebesgue measure on $Bbb R$ and the Cantor set in the interval $[0,1]$ is an example.
My guess is that the same argument can be repeated for $mu_F$, if there is an interval $[a,b]$ where $F$ is continuous (hence making sure that any singleton set has measure 0), and construct a "Cantor set" there.
But an increasing function can have a dense set of discontinuities as shown in here.
What can one do in this case?
measure-theory
asked Dec 21 '18 at 6:02
UserAUserA
540216
$endgroup$
add a comment |
$begingroup$
Let $mu_F$ be the Lebesgue-Stieltjes measure on $Bbb R$ associated with the increasing function $F:Bbb Rto Bbb R$. Construct an uncountable set of measure 0 for $mu_F$.
When $F(x)=x$ and we get the Lebesgue measure on $Bbb R$ and the Cantor set in the interval $[0,1]$ is an example.
My guess is that the same argument can be repeated for $mu_F$, if there is an interval $[a,b]$ where $F$ is continuous (hence making sure that any singleton set has measure 0), and construct a "Cantor set" there.
But an increasing function can have a dense set of discontinuities as shown in here.
What can one do in this case?
measure-theory
asked Dec 21 '18 at 6:02
UserAUserA
540216
$endgroup$
add a comment |
$begingroup$
Let $mu_F$ be the Lebesgue-Stieltjes measure on $Bbb R$ associated with the increasing function $F:Bbb Rto Bbb R$. Construct an uncountable set of measure 0 for $mu_F$.
When $F(x)=x$ and we get the Lebesgue measure on $Bbb R$ and the Cantor set in the interval $[0,1]$ is an example.
My guess is that the same argument can be repeated for $mu_F$, if there is an interval $[a,b]$ where $F$ is continuous (hence making sure that any singleton set has measure 0), and construct a "Cantor set" there.
But an increasing function can have a dense set of discontinuities as shown in here.
What can one do in this case?
measure-theory
asked Dec 21 '18 at 6:02
UserAUserA
540216
$endgroup$
Let $mu_F$ be the Lebesgue-Stieltjes measure on $Bbb R$ associated with the increasing function $F:Bbb Rto Bbb R$. Construct an uncountable set of measure 0 for $mu_F$.
When $F(x)=x$ and we get the Lebesgue measure on $Bbb R$ and the Cantor set in the interval $[0,1]$ is an example.
My guess is that the same argument can be repeated for $mu_F$, if there is an interval $[a,b]$ where $F$ is continuous (hence making sure that any singleton set has measure 0), and construct a "Cantor set" there.
But an increasing function can have a dense set of discontinuities as shown in here.
What can one do in this case?
measure-theory
measure-theory
asked Dec 21 '18 at 6:02
UserAUserA
540216
asked Dec 21 '18 at 6:02
UserAUserA
540216
asked Dec 21 '18 at 6:02
UserAUserA
540216
asked Dec 21 '18 at 6:02
UserAUserA
540216
asked Dec 21 '18 at 6:02
UserAUserA
540216
540216
add a comment |
add a comment |
1 Answer
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Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $mu_F=lambda+nu$ where $lambdaperp m$ on some set $E$ with $m(E^c)=0=lambda(E)$ and $null m$. By regularity, there exists a compact $Ksubset E$ with $m(K)>0$. Define $f(x)=m(Kcap (-infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(Kcap (x,y])le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=frac{m(K)}3$ and $f(b)=frac{2m(K)}3$, in particular, $m(Kcap(-infty,a])=m(Kcap[b,infty))=frac{m(K)}3$. Thus $Kcap (-infty,a]$ and $Kcap [b,infty)$ are compact with positive measure. Repeating this process inductively yields a $Csubset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $mu_F(C)le lambda(E)+nu(C)=0$.
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $mu_F=lambda+nu$ where $lambdaperp m$ on some set $E$ with $m(E^c)=0=lambda(E)$ and $null m$. By regularity, there exists a compact $Ksubset E$ with $m(K)>0$. Define $f(x)=m(Kcap (-infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(Kcap (x,y])le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=frac{m(K)}3$ and $f(b)=frac{2m(K)}3$, in particular, $m(Kcap(-infty,a])=m(Kcap[b,infty))=frac{m(K)}3$. Thus $Kcap (-infty,a]$ and $Kcap [b,infty)$ are compact with positive measure. Repeating this process inductively yields a $Csubset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $mu_F(C)le lambda(E)+nu(C)=0$.
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
add a comment |
$begingroup$
Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $mu_F=lambda+nu$ where $lambdaperp m$ on some set $E$ with $m(E^c)=0=lambda(E)$ and $null m$. By regularity, there exists a compact $Ksubset E$ with $m(K)>0$. Define $f(x)=m(Kcap (-infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(Kcap (x,y])le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=frac{m(K)}3$ and $f(b)=frac{2m(K)}3$, in particular, $m(Kcap(-infty,a])=m(Kcap[b,infty))=frac{m(K)}3$. Thus $Kcap (-infty,a]$ and $Kcap [b,infty)$ are compact with positive measure. Repeating this process inductively yields a $Csubset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $mu_F(C)le lambda(E)+nu(C)=0$.
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
add a comment |
$begingroup$
Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $mu_F=lambda+nu$ where $lambdaperp m$ on some set $E$ with $m(E^c)=0=lambda(E)$ and $null m$. By regularity, there exists a compact $Ksubset E$ with $m(K)>0$. Define $f(x)=m(Kcap (-infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(Kcap (x,y])le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=frac{m(K)}3$ and $f(b)=frac{2m(K)}3$, in particular, $m(Kcap(-infty,a])=m(Kcap[b,infty))=frac{m(K)}3$. Thus $Kcap (-infty,a]$ and $Kcap [b,infty)$ are compact with positive measure. Repeating this process inductively yields a $Csubset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $mu_F(C)le lambda(E)+nu(C)=0$.
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
$endgroup$
Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $mu_F=lambda+nu$ where $lambdaperp m$ on some set $E$ with $m(E^c)=0=lambda(E)$ and $null m$. By regularity, there exists a compact $Ksubset E$ with $m(K)>0$. Define $f(x)=m(Kcap (-infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(Kcap (x,y])le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=frac{m(K)}3$ and $f(b)=frac{2m(K)}3$, in particular, $m(Kcap(-infty,a])=m(Kcap[b,infty))=frac{m(K)}3$. Thus $Kcap (-infty,a]$ and $Kcap [b,infty)$ are compact with positive measure. Repeating this process inductively yields a $Csubset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $mu_F(C)le lambda(E)+nu(C)=0$.
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
answered Dec 23 '18 at 4:59
Guacho PerezGuacho Perez
3,92911132
3,92911132
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
add a comment |
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
$begingroup$
(+1) In particular, $lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $nu$ is the continuous part. For any Borel set $A in mathcal{B}(mathbb{R})$ we have $mu_F(A) = m^ast(F(A)) + underbrace{sumlimits_{p in A cap D} mu_F({p})}_{ = lambda(A)}$.
$endgroup$
– r9m
Dec 23 '18 at 7:14
add a comment |
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