measure preserving transformation inequality
Let $(lambda,Lambda)$ denote the Lebesgue measure space on $(0,infty)$. Suppose $E$ and $F$ are Lebesgue-measurable subsets of $(0,infty)$. A map $m:Fto E$ is called measure-presering whenever $lambda(m^{-1}(A))=lambda(A)$ for each measurable subset $A$ of $E$. It is called order-preserving if $aleq b$ in $F$ implies $m(a)leq m(b)$ in $E$. Denote by $mathbb{MO}(F,E)$ the set of all bijective maps $m:Fto E$ such that $m$ and $m^{-1}$ are both order-preserving and measure-preserving.
Fix $rin(0,infty)$. Let $W:(0,infty)to(0,infty)$ be a positive decreasing function and $f:(0,infty)to[0,infty)$ be a nonnegative function increasing on $(0,r]$ and zero on $(r,infty)$.
Conjecture. Suppose $Esubset(0,r]$ with $b=lambda(E)=lambda(F)leq r$, and $minmathbb{MO}(F,E)$. Then
$$int_0^inftylambda{xin(0,b]:f(t+r-b)W(t)>x};dtgeqint_0^inftylambda{xin F:(fcirc m)(t)W(t)>x};dt.$$
Or, equivalently,
$$int_0^bf(t+r-b)W(t);dtgeqint_F(fcirc m)(t)W(t);dt.$$
This seems intuitively obvious, and would be easy to prove if we were working in $mathbb{N}$ endowed with the counting measure instead of $(0,infty)$ endowed with the Lebesgue measure. Unfortunately, Lebesgue-measurable sets can be very ugly indeed, and so that makes working with them difficult sometimes.
The following may help.
Fact. There exists a surjection $tau:Eto[0,b]$ which is both measure-preserving and order-preserving. (However, it is not necessarily injective, and hence need not be invertible.) A similar map exists for $F$.
real-analysis measure-theory
asked Nov 19 '18 at 0:14
Ben W
1,423513
add a comment |
Let $(lambda,Lambda)$ denote the Lebesgue measure space on $(0,infty)$. Suppose $E$ and $F$ are Lebesgue-measurable subsets of $(0,infty)$. A map $m:Fto E$ is called measure-presering whenever $lambda(m^{-1}(A))=lambda(A)$ for each measurable subset $A$ of $E$. It is called order-preserving if $aleq b$ in $F$ implies $m(a)leq m(b)$ in $E$. Denote by $mathbb{MO}(F,E)$ the set of all bijective maps $m:Fto E$ such that $m$ and $m^{-1}$ are both order-preserving and measure-preserving.
Fix $rin(0,infty)$. Let $W:(0,infty)to(0,infty)$ be a positive decreasing function and $f:(0,infty)to[0,infty)$ be a nonnegative function increasing on $(0,r]$ and zero on $(r,infty)$.
Conjecture. Suppose $Esubset(0,r]$ with $b=lambda(E)=lambda(F)leq r$, and $minmathbb{MO}(F,E)$. Then
$$int_0^inftylambda{xin(0,b]:f(t+r-b)W(t)>x};dtgeqint_0^inftylambda{xin F:(fcirc m)(t)W(t)>x};dt.$$
Or, equivalently,
$$int_0^bf(t+r-b)W(t);dtgeqint_F(fcirc m)(t)W(t);dt.$$
This seems intuitively obvious, and would be easy to prove if we were working in $mathbb{N}$ endowed with the counting measure instead of $(0,infty)$ endowed with the Lebesgue measure. Unfortunately, Lebesgue-measurable sets can be very ugly indeed, and so that makes working with them difficult sometimes.
The following may help.
Fact. There exists a surjection $tau:Eto[0,b]$ which is both measure-preserving and order-preserving. (However, it is not necessarily injective, and hence need not be invertible.) A similar map exists for $F$.
real-analysis measure-theory
asked Nov 19 '18 at 0:14
Ben W
1,423513
add a comment |
Let $(lambda,Lambda)$ denote the Lebesgue measure space on $(0,infty)$. Suppose $E$ and $F$ are Lebesgue-measurable subsets of $(0,infty)$. A map $m:Fto E$ is called measure-presering whenever $lambda(m^{-1}(A))=lambda(A)$ for each measurable subset $A$ of $E$. It is called order-preserving if $aleq b$ in $F$ implies $m(a)leq m(b)$ in $E$. Denote by $mathbb{MO}(F,E)$ the set of all bijective maps $m:Fto E$ such that $m$ and $m^{-1}$ are both order-preserving and measure-preserving.
Fix $rin(0,infty)$. Let $W:(0,infty)to(0,infty)$ be a positive decreasing function and $f:(0,infty)to[0,infty)$ be a nonnegative function increasing on $(0,r]$ and zero on $(r,infty)$.
Conjecture. Suppose $Esubset(0,r]$ with $b=lambda(E)=lambda(F)leq r$, and $minmathbb{MO}(F,E)$. Then
$$int_0^inftylambda{xin(0,b]:f(t+r-b)W(t)>x};dtgeqint_0^inftylambda{xin F:(fcirc m)(t)W(t)>x};dt.$$
Or, equivalently,
$$int_0^bf(t+r-b)W(t);dtgeqint_F(fcirc m)(t)W(t);dt.$$
This seems intuitively obvious, and would be easy to prove if we were working in $mathbb{N}$ endowed with the counting measure instead of $(0,infty)$ endowed with the Lebesgue measure. Unfortunately, Lebesgue-measurable sets can be very ugly indeed, and so that makes working with them difficult sometimes.
The following may help.
Fact. There exists a surjection $tau:Eto[0,b]$ which is both measure-preserving and order-preserving. (However, it is not necessarily injective, and hence need not be invertible.) A similar map exists for $F$.
real-analysis measure-theory
asked Nov 19 '18 at 0:14
Ben W
1,423513
Let $(lambda,Lambda)$ denote the Lebesgue measure space on $(0,infty)$. Suppose $E$ and $F$ are Lebesgue-measurable subsets of $(0,infty)$. A map $m:Fto E$ is called measure-presering whenever $lambda(m^{-1}(A))=lambda(A)$ for each measurable subset $A$ of $E$. It is called order-preserving if $aleq b$ in $F$ implies $m(a)leq m(b)$ in $E$. Denote by $mathbb{MO}(F,E)$ the set of all bijective maps $m:Fto E$ such that $m$ and $m^{-1}$ are both order-preserving and measure-preserving.
Fix $rin(0,infty)$. Let $W:(0,infty)to(0,infty)$ be a positive decreasing function and $f:(0,infty)to[0,infty)$ be a nonnegative function increasing on $(0,r]$ and zero on $(r,infty)$.
Conjecture. Suppose $Esubset(0,r]$ with $b=lambda(E)=lambda(F)leq r$, and $minmathbb{MO}(F,E)$. Then
$$int_0^inftylambda{xin(0,b]:f(t+r-b)W(t)>x};dtgeqint_0^inftylambda{xin F:(fcirc m)(t)W(t)>x};dt.$$
Or, equivalently,
$$int_0^bf(t+r-b)W(t);dtgeqint_F(fcirc m)(t)W(t);dt.$$
This seems intuitively obvious, and would be easy to prove if we were working in $mathbb{N}$ endowed with the counting measure instead of $(0,infty)$ endowed with the Lebesgue measure. Unfortunately, Lebesgue-measurable sets can be very ugly indeed, and so that makes working with them difficult sometimes.
The following may help.
Fact. There exists a surjection $tau:Eto[0,b]$ which is both measure-preserving and order-preserving. (However, it is not necessarily injective, and hence need not be invertible.) A similar map exists for $F$.
real-analysis measure-theory
real-analysis measure-theory
asked Nov 19 '18 at 0:14
Ben W
1,423513
asked Nov 19 '18 at 0:14
Ben W
1,423513
edited Nov 19 '18 at 13:32
asked Nov 19 '18 at 0:14
Ben W
1,423513
asked Nov 19 '18 at 0:14
Ben W
1,423513
asked Nov 19 '18 at 0:14
Ben W
1,423513
1,423513
add a comment |
add a comment |
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