Lebesgue decomposition of a measure
up vote
2
down vote
favorite
I have a similar question as here.
Lesbesgue-Stieltjes measure $mu_F$ with the function $F=lvert xrvertlfloor xrfloor$. i.e. $mu_F([a,b))=F(b)-F(a)$. I am trying to find the Lebesgue-Radon-Nikodym decomposition of $mu_F$ with respect to the Lebesgue measure $m$.
Now, I know this function has countably many discontinuities at each integer-valued $x$, so I believe I want the mutually singular part to be the Dirac Measure of each integer, and take the absolute continuous part to be the Lebesgue measure itself. But it would have to be a countable sum of Dirac measures for the discontinuity points… and I am not sure if I am allowed to do that, or if my approach is ok.
I would love to get some feedback/a point in the right direction or any help in general.
real-analysis measure-theory
edited Nov 23 at 7:04
user302797
19.7k92252
asked Nov 21 at 3:40
Mog
549
|
show 1 more comment
up vote
2
down vote
favorite
I have a similar question as here.
Lesbesgue-Stieltjes measure $mu_F$ with the function $F=lvert xrvertlfloor xrfloor$. i.e. $mu_F([a,b))=F(b)-F(a)$. I am trying to find the Lebesgue-Radon-Nikodym decomposition of $mu_F$ with respect to the Lebesgue measure $m$.
Now, I know this function has countably many discontinuities at each integer-valued $x$, so I believe I want the mutually singular part to be the Dirac Measure of each integer, and take the absolute continuous part to be the Lebesgue measure itself. But it would have to be a countable sum of Dirac measures for the discontinuity points… and I am not sure if I am allowed to do that, or if my approach is ok.
I would love to get some feedback/a point in the right direction or any help in general.
real-analysis measure-theory
edited Nov 23 at 7:04
user302797
19.7k92252
asked Nov 21 at 3:40
Mog
549
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a similar question as here.
Lesbesgue-Stieltjes measure $mu_F$ with the function $F=lvert xrvertlfloor xrfloor$. i.e. $mu_F([a,b))=F(b)-F(a)$. I am trying to find the Lebesgue-Radon-Nikodym decomposition of $mu_F$ with respect to the Lebesgue measure $m$.
Now, I know this function has countably many discontinuities at each integer-valued $x$, so I believe I want the mutually singular part to be the Dirac Measure of each integer, and take the absolute continuous part to be the Lebesgue measure itself. But it would have to be a countable sum of Dirac measures for the discontinuity points… and I am not sure if I am allowed to do that, or if my approach is ok.
I would love to get some feedback/a point in the right direction or any help in general.
real-analysis measure-theory
edited Nov 23 at 7:04
user302797
19.7k92252
asked Nov 21 at 3:40
Mog
549
I have a similar question as here.
Lesbesgue-Stieltjes measure $mu_F$ with the function $F=lvert xrvertlfloor xrfloor$. i.e. $mu_F([a,b))=F(b)-F(a)$. I am trying to find the Lebesgue-Radon-Nikodym decomposition of $mu_F$ with respect to the Lebesgue measure $m$.
Now, I know this function has countably many discontinuities at each integer-valued $x$, so I believe I want the mutually singular part to be the Dirac Measure of each integer, and take the absolute continuous part to be the Lebesgue measure itself. But it would have to be a countable sum of Dirac measures for the discontinuity points… and I am not sure if I am allowed to do that, or if my approach is ok.
I would love to get some feedback/a point in the right direction or any help in general.
real-analysis measure-theory
real-analysis measure-theory
edited Nov 23 at 7:04
user302797
19.7k92252
asked Nov 21 at 3:40
Mog
549
edited Nov 23 at 7:04
user302797
19.7k92252
asked Nov 21 at 3:40
Mog
549
edited Nov 23 at 7:04
user302797
19.7k92252
edited Nov 23 at 7:04
user302797
19.7k92252
edited Nov 23 at 7:04
user302797
19.7k92252
19.7k92252
asked Nov 21 at 3:40
Mog
549
asked Nov 21 at 3:40
Mog
549
asked Nov 21 at 3:40
Mog
549
549
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22
|
show 1 more comment
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.
For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.
To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.
answered Nov 23 at 7:02
user302797
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007229%2flebesgue-decomposition-of-a-measure%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.
For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.
To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.
answered Nov 23 at 7:02
user302797
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
add a comment |
up vote
3
down vote
accepted
$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.
For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.
To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.
answered Nov 23 at 7:02
user302797
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.
For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.
To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.
answered Nov 23 at 7:02
user302797
19.7k92252
$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.
For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.
To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.
answered Nov 23 at 7:02
user302797
19.7k92252
answered Nov 23 at 7:02
user302797
19.7k92252
answered Nov 23 at 7:02
user302797
19.7k92252
answered Nov 23 at 7:02
user302797
19.7k92252
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
add a comment |
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
1
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– user302797
Nov 23 at 8:58
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007229%2flebesgue-decomposition-of-a-measure%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Each integer is an atom, and its size is $|x|.$ I think there is no diffuse part that is continuous and the absolutely continuous part will be obtained by difference.
– Will M.
Nov 21 at 4:07
@WillM. What do you mean by "there is no diffuse part"? And do you mean the absolute continuous part with respect to Lebesgue measure is simply difference of $F$ at endpoints, or do you mean something else? Also, wouldn't the atom's size be 1?
– Mog
Nov 21 at 4:09
On the one hand, a measure $mu$ is "diffuse" if $mu(mathrm{K}) = 0$ for all finite $mathrm{K}.$ On the other hand, you will construct a measure $mu$ that is atomic (the atoms of $mu_F$) then the absolutely continuous part, my guess is, $mu_F-mu.$
– Will M.
Nov 21 at 4:12
@WillM. I'm not sure if I follow.. If you have the time, do you mind writing out the answer?
– Mog
Nov 21 at 4:20
This is a really interesting question. If you could briefly refresh my memory what does it mean to take a measure with respect to a function?
– The Great Duck
Nov 23 at 6:22