Mixing
Lebesgue decomposition of a measure
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 '18 at 7:04
Saad
19.7k92252
asked Nov 21 '18 at 3:40
Mog
549
|
show 1 more comment
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 '18 at 7:04
Saad
19.7k92252
asked Nov 21 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 6:22
|
show 1 more comment
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 '18 at 7:04
Saad
19.7k92252
asked Nov 21 '18 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 '18 at 7:04
Saad
19.7k92252
asked Nov 21 '18 at 3:40
Mog
549
edited Nov 23 '18 at 7:04
Saad
19.7k92252
asked Nov 21 '18 at 3:40
Mog
549
edited Nov 23 '18 at 7:04
Saad
19.7k92252
edited Nov 23 '18 at 7:04
Saad
19.7k92252
edited Nov 23 '18 at 7:04
Saad
19.7k92252
19.7k92252
asked Nov 21 '18 at 3:40
Mog
549
asked Nov 21 '18 at 3:40
Mog
549
asked Nov 21 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 6:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
$defd{mathrm{d}}$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 '18 at 7:02
Saad
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 at 8:58
add a comment |
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1 Answer
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$defd{mathrm{d}}$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 '18 at 7:02
Saad
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 at 8:58
add a comment |
$defd{mathrm{d}}$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 '18 at 7:02
Saad
19.7k92252
1
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 at 8:58
add a comment |
$defd{mathrm{d}}$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 '18 at 7:02
Saad
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 '18 at 7:02
Saad
19.7k92252
answered Nov 23 '18 at 7:02
Saad
19.7k92252
answered Nov 23 '18 at 7:02
Saad
19.7k92252
answered Nov 23 '18 at 7:02
Saad
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 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 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 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 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 '18 at 7:08
Thank you for your answer. What do the $dx$'s next to $mu_c, mu_s$ mean?
– Mog
Nov 23 '18 at 7:08
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 at 8:58
@Mog Simply a notation as in $μ(A)=int_Aμ(mathrm dx)$.
– Saad
Nov 23 '18 at 8:58
add a comment |
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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 '18 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 '18 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 '18 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 '18 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 '18 at 6:22