Check if $supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/kright}to 0$
Multi tool use
Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set
$$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$
I want to see if it is true that $(R_k)to 0$.
What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
$$
lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
$$
which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.
Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that
$$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$
and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.
Two questions here:
It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.
If the statement of the title is true, there is a simpler way to show the same result?
analysis measure-theory proof-verification lebesgue-integral alternative-proof
add a comment |
Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set
$$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$
I want to see if it is true that $(R_k)to 0$.
What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
$$
lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
$$
which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.
Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that
$$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$
and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.
Two questions here:
It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.
If the statement of the title is true, there is a simpler way to show the same result?
analysis measure-theory proof-verification lebesgue-integral alternative-proof
add a comment |
Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set
$$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$
I want to see if it is true that $(R_k)to 0$.
What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
$$
lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
$$
which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.
Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that
$$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$
and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.
Two questions here:
It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.
If the statement of the title is true, there is a simpler way to show the same result?
analysis measure-theory proof-verification lebesgue-integral alternative-proof
Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set
$$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$
I want to see if it is true that $(R_k)to 0$.
What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
$$
lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
$$
which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.
Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that
$$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$
and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.
Two questions here:
It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.
If the statement of the title is true, there is a simpler way to show the same result?
analysis measure-theory proof-verification lebesgue-integral alternative-proof
analysis measure-theory proof-verification lebesgue-integral alternative-proof
edited Dec 9 at 14:19
Did
246k23220454
246k23220454
asked Dec 9 at 7:52
Masacroso
12.9k41746
12.9k41746
add a comment |
add a comment |
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Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.
The statement is equivalent to the absolute continuity of the finite measure
$$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
$$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
begin{align}
int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
& le varepsilon/2 + K lambda(A) < varepsilon.
end{align}
In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
add a comment |
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Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.
The statement is equivalent to the absolute continuity of the finite measure
$$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
$$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
begin{align}
int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
& le varepsilon/2 + K lambda(A) < varepsilon.
end{align}
In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
add a comment |
Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.
The statement is equivalent to the absolute continuity of the finite measure
$$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
$$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
begin{align}
int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
& le varepsilon/2 + K lambda(A) < varepsilon.
end{align}
In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
add a comment |
Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.
The statement is equivalent to the absolute continuity of the finite measure
$$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
$$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
begin{align}
int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
& le varepsilon/2 + K lambda(A) < varepsilon.
end{align}
In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.
Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.
The statement is equivalent to the absolute continuity of the finite measure
$$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
$$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
begin{align}
int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
& le varepsilon/2 + K lambda(A) < varepsilon.
end{align}
In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.
edited Dec 9 at 20:27
answered Dec 9 at 13:25
p4sch
4,760217
4,760217
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
add a comment |
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
– Masacroso
Dec 9 at 19:39
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Ah, okay! However, my argumentation is much easier. :-)
– p4sch
Dec 9 at 20:02
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
Yes, of course! Typo corrected. :-)
– p4sch
Dec 9 at 20:28
add a comment |
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