Lusin's theorem on arbitrary measure space
$begingroup$
This question is generalization of Lusin's theorem, though it appears long.
Let $(X, rho)$ and $(Y, sigma)$ be metric spaces with $(Y, sigma)$ separable, and let $mu$ be a finite Borel measure on $X$. Let $f : X to Y$ be $mathcal B_X, mathcal B_Y)$-measurable and let $epsilon > 0$. Prove that there is a closed set $F subset X$ such that $mu(F^c) < epsilon$ and $f|_F$ is continuous.
Here, $mathcal B_X$ denote the Borel $sigma$ algwbra on $X$, and similarly for $mathcal B_Y$. I have proven that $mathcal B_Y$ is generated by the collection of all open balls in $Y$.
I am to model after a version of Lusin's theorem that uses regularity of finite measures, as the following:
"The real line has only countably many open intervals whose end-points are both rational. Let ${ I_n }$ be an enumeration of them.
For each $n$, the regularity of $µ$ gives closed subsets $A_n ⊂ f^{-1}[I_n]$ and $B_n ⊂ f^{-1}[mathbb R - I_n]$ such that $$µ((A_n ∪ B_n)^c) = µ(f^{-1}[I_n] - A_n) + µ(f^{-1}[mathbb R - I_n] - B_n) < ε2^{-n}.$$
Let $F := bigcap_n(A_n ∪ B_n)$. This is closed, and it satisfies
$$mu(F^c) = µ( bigcup (A_n ∪ B_n)^c) ≤ sum µ((A_n ∪ B_n)^c) < ε.$$
Let us show that $f|F$ is continuous. Let $x ∈ F$ and $ε > 0.$ By the density of $mathbb Q$ in $ mathbb R$, there is a rational open interval In which satisfies $f(x) ∈ I_n ⊂ (f(x) − ε, f(x) + ε)$.Thus $ x$ lies in both $F$ and $f^{-1}[I_n]$, and so $x ∈ A_n ⊂ B_n^c.$ Since $B_n$ is closed, we can now choose $ δ > 0$ so that $B_δ(x)$ is disjoint from $B_n$. Therefore, if $y ∈ F$ satisfies that $ρ(x, y) < δ$, then it cannot lie in $B_n$, so it must also lie in $A_n$. This implies that $f(y) ∈ I_n ⊂ (f(x) − ε, f(x) + ε).$
In adapting this proof we assume that for a given borel set $I$ of $Y$ (which, by what I have proved, is an open ball- and specifically, an open ball of rational radius) there is a closed set $K$ and an open set $U$ such that $K subset I subset U$, with all sets concentric balls and $K$ and $U$ with radius close enough to $I$ such that $mu(U-K)$ small.
And then proceed similarly as above.
Is my reasoning correct?
measure-theory proof-verification borel-measures
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
$endgroup$
add a comment |
$begingroup$
This question is generalization of Lusin's theorem, though it appears long.
Let $(X, rho)$ and $(Y, sigma)$ be metric spaces with $(Y, sigma)$ separable, and let $mu$ be a finite Borel measure on $X$. Let $f : X to Y$ be $mathcal B_X, mathcal B_Y)$-measurable and let $epsilon > 0$. Prove that there is a closed set $F subset X$ such that $mu(F^c) < epsilon$ and $f|_F$ is continuous.
Here, $mathcal B_X$ denote the Borel $sigma$ algwbra on $X$, and similarly for $mathcal B_Y$. I have proven that $mathcal B_Y$ is generated by the collection of all open balls in $Y$.
I am to model after a version of Lusin's theorem that uses regularity of finite measures, as the following:
"The real line has only countably many open intervals whose end-points are both rational. Let ${ I_n }$ be an enumeration of them.
For each $n$, the regularity of $µ$ gives closed subsets $A_n ⊂ f^{-1}[I_n]$ and $B_n ⊂ f^{-1}[mathbb R - I_n]$ such that $$µ((A_n ∪ B_n)^c) = µ(f^{-1}[I_n] - A_n) + µ(f^{-1}[mathbb R - I_n] - B_n) < ε2^{-n}.$$
Let $F := bigcap_n(A_n ∪ B_n)$. This is closed, and it satisfies
$$mu(F^c) = µ( bigcup (A_n ∪ B_n)^c) ≤ sum µ((A_n ∪ B_n)^c) < ε.$$
Let us show that $f|F$ is continuous. Let $x ∈ F$ and $ε > 0.$ By the density of $mathbb Q$ in $ mathbb R$, there is a rational open interval In which satisfies $f(x) ∈ I_n ⊂ (f(x) − ε, f(x) + ε)$.Thus $ x$ lies in both $F$ and $f^{-1}[I_n]$, and so $x ∈ A_n ⊂ B_n^c.$ Since $B_n$ is closed, we can now choose $ δ > 0$ so that $B_δ(x)$ is disjoint from $B_n$. Therefore, if $y ∈ F$ satisfies that $ρ(x, y) < δ$, then it cannot lie in $B_n$, so it must also lie in $A_n$. This implies that $f(y) ∈ I_n ⊂ (f(x) − ε, f(x) + ε).$
In adapting this proof we assume that for a given borel set $I$ of $Y$ (which, by what I have proved, is an open ball- and specifically, an open ball of rational radius) there is a closed set $K$ and an open set $U$ such that $K subset I subset U$, with all sets concentric balls and $K$ and $U$ with radius close enough to $I$ such that $mu(U-K)$ small.
And then proceed similarly as above.
Is my reasoning correct?
measure-theory proof-verification borel-measures
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
$endgroup$
$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02
add a comment |
$begingroup$
This question is generalization of Lusin's theorem, though it appears long.
Let $(X, rho)$ and $(Y, sigma)$ be metric spaces with $(Y, sigma)$ separable, and let $mu$ be a finite Borel measure on $X$. Let $f : X to Y$ be $mathcal B_X, mathcal B_Y)$-measurable and let $epsilon > 0$. Prove that there is a closed set $F subset X$ such that $mu(F^c) < epsilon$ and $f|_F$ is continuous.
Here, $mathcal B_X$ denote the Borel $sigma$ algwbra on $X$, and similarly for $mathcal B_Y$. I have proven that $mathcal B_Y$ is generated by the collection of all open balls in $Y$.
I am to model after a version of Lusin's theorem that uses regularity of finite measures, as the following:
"The real line has only countably many open intervals whose end-points are both rational. Let ${ I_n }$ be an enumeration of them.
For each $n$, the regularity of $µ$ gives closed subsets $A_n ⊂ f^{-1}[I_n]$ and $B_n ⊂ f^{-1}[mathbb R - I_n]$ such that $$µ((A_n ∪ B_n)^c) = µ(f^{-1}[I_n] - A_n) + µ(f^{-1}[mathbb R - I_n] - B_n) < ε2^{-n}.$$
Let $F := bigcap_n(A_n ∪ B_n)$. This is closed, and it satisfies
$$mu(F^c) = µ( bigcup (A_n ∪ B_n)^c) ≤ sum µ((A_n ∪ B_n)^c) < ε.$$
Let us show that $f|F$ is continuous. Let $x ∈ F$ and $ε > 0.$ By the density of $mathbb Q$ in $ mathbb R$, there is a rational open interval In which satisfies $f(x) ∈ I_n ⊂ (f(x) − ε, f(x) + ε)$.Thus $ x$ lies in both $F$ and $f^{-1}[I_n]$, and so $x ∈ A_n ⊂ B_n^c.$ Since $B_n$ is closed, we can now choose $ δ > 0$ so that $B_δ(x)$ is disjoint from $B_n$. Therefore, if $y ∈ F$ satisfies that $ρ(x, y) < δ$, then it cannot lie in $B_n$, so it must also lie in $A_n$. This implies that $f(y) ∈ I_n ⊂ (f(x) − ε, f(x) + ε).$
In adapting this proof we assume that for a given borel set $I$ of $Y$ (which, by what I have proved, is an open ball- and specifically, an open ball of rational radius) there is a closed set $K$ and an open set $U$ such that $K subset I subset U$, with all sets concentric balls and $K$ and $U$ with radius close enough to $I$ such that $mu(U-K)$ small.
And then proceed similarly as above.
Is my reasoning correct?
measure-theory proof-verification borel-measures
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
$endgroup$
This question is generalization of Lusin's theorem, though it appears long.
Let $(X, rho)$ and $(Y, sigma)$ be metric spaces with $(Y, sigma)$ separable, and let $mu$ be a finite Borel measure on $X$. Let $f : X to Y$ be $mathcal B_X, mathcal B_Y)$-measurable and let $epsilon > 0$. Prove that there is a closed set $F subset X$ such that $mu(F^c) < epsilon$ and $f|_F$ is continuous.
Here, $mathcal B_X$ denote the Borel $sigma$ algwbra on $X$, and similarly for $mathcal B_Y$. I have proven that $mathcal B_Y$ is generated by the collection of all open balls in $Y$.
I am to model after a version of Lusin's theorem that uses regularity of finite measures, as the following:
"The real line has only countably many open intervals whose end-points are both rational. Let ${ I_n }$ be an enumeration of them.
For each $n$, the regularity of $µ$ gives closed subsets $A_n ⊂ f^{-1}[I_n]$ and $B_n ⊂ f^{-1}[mathbb R - I_n]$ such that $$µ((A_n ∪ B_n)^c) = µ(f^{-1}[I_n] - A_n) + µ(f^{-1}[mathbb R - I_n] - B_n) < ε2^{-n}.$$
Let $F := bigcap_n(A_n ∪ B_n)$. This is closed, and it satisfies
$$mu(F^c) = µ( bigcup (A_n ∪ B_n)^c) ≤ sum µ((A_n ∪ B_n)^c) < ε.$$
Let us show that $f|F$ is continuous. Let $x ∈ F$ and $ε > 0.$ By the density of $mathbb Q$ in $ mathbb R$, there is a rational open interval In which satisfies $f(x) ∈ I_n ⊂ (f(x) − ε, f(x) + ε)$.Thus $ x$ lies in both $F$ and $f^{-1}[I_n]$, and so $x ∈ A_n ⊂ B_n^c.$ Since $B_n$ is closed, we can now choose $ δ > 0$ so that $B_δ(x)$ is disjoint from $B_n$. Therefore, if $y ∈ F$ satisfies that $ρ(x, y) < δ$, then it cannot lie in $B_n$, so it must also lie in $A_n$. This implies that $f(y) ∈ I_n ⊂ (f(x) − ε, f(x) + ε).$
In adapting this proof we assume that for a given borel set $I$ of $Y$ (which, by what I have proved, is an open ball- and specifically, an open ball of rational radius) there is a closed set $K$ and an open set $U$ such that $K subset I subset U$, with all sets concentric balls and $K$ and $U$ with radius close enough to $I$ such that $mu(U-K)$ small.
And then proceed similarly as above.
Is my reasoning correct?
measure-theory proof-verification borel-measures
measure-theory proof-verification borel-measures
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
edited Nov 25 '18 at 9:19
Cute Brownie
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
asked Nov 25 '18 at 5:27
Cute BrownieCute Brownie
990416
990416
$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02
add a comment |
$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02
$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02
add a comment |
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$begingroup$
I am to model after a version of Lusin's theorem that uses regularity of finite measures --- Perhaps you could post the "a version" you speak of, and begin the task of modifying it to your situation, and then point out where things seem to break down for you (or where you start getting confused as to how to proceed).
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 8:30
$begingroup$
@DaveL.Renfro thanks, I indicated which part of the model proff I don't get.
$endgroup$
– Cute Brownie
Nov 25 '18 at 9:12
$begingroup$
I saw that, but unfortunately I've been really busy with work lately (stuff I get paid to do) and don't have time to look into this anytime in the next few days. However, I thought by giving the version you want to start with (as I mentioned in my other comment) that you would generate some responses from others. I'm surprised that you haven't, by the way.
$endgroup$
– Dave L. Renfro
Nov 25 '18 at 19:02