Prove that $f$ is integrable if and only if $sum^infty_{n=1} mu({x in X : f(x) ge n}) < infty$
$begingroup$
Problem statement: Suppose that $mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $sum^infty_{n=1} mu({x in X : f(x) ge n}) < infty$.
My attempt at a solution: Let $A_n = {x in X : f(x) ge n}$. To show that $sum^infty_{n=1} mu ({x in X : f(x) ge n}) < infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x in X$ belong to at most finitely many $A_n$. Thus, the set ${x in X : f(x) = infty}$ has measure $0$. Now, this, together with the fact that $mu(X) < infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) le M$ for all $x$.
For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $sum^m_{n=1}mu(A_n)$, is bounded, but I'm not sure how to do so.
real-analysis analysis lebesgue-integral lebesgue-measure
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
$endgroup$
add a comment |
$begingroup$
Problem statement: Suppose that $mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $sum^infty_{n=1} mu({x in X : f(x) ge n}) < infty$.
My attempt at a solution: Let $A_n = {x in X : f(x) ge n}$. To show that $sum^infty_{n=1} mu ({x in X : f(x) ge n}) < infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x in X$ belong to at most finitely many $A_n$. Thus, the set ${x in X : f(x) = infty}$ has measure $0$. Now, this, together with the fact that $mu(X) < infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) le M$ for all $x$.
For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $sum^m_{n=1}mu(A_n)$, is bounded, but I'm not sure how to do so.
real-analysis analysis lebesgue-integral lebesgue-measure
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
$endgroup$
$begingroup$
What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
$endgroup$
– user2566092
Aug 6 '15 at 19:45
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@user2566092 yes, sorry, totally screwed up there! it's been edited
$endgroup$
– poppy3345
Aug 6 '15 at 19:46
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@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
$endgroup$
– Gary.
Aug 6 '15 at 20:48
add a comment |
$begingroup$
Problem statement: Suppose that $mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $sum^infty_{n=1} mu({x in X : f(x) ge n}) < infty$.
My attempt at a solution: Let $A_n = {x in X : f(x) ge n}$. To show that $sum^infty_{n=1} mu ({x in X : f(x) ge n}) < infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x in X$ belong to at most finitely many $A_n$. Thus, the set ${x in X : f(x) = infty}$ has measure $0$. Now, this, together with the fact that $mu(X) < infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) le M$ for all $x$.
For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $sum^m_{n=1}mu(A_n)$, is bounded, but I'm not sure how to do so.
real-analysis analysis lebesgue-integral lebesgue-measure
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
$endgroup$
Problem statement: Suppose that $mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $sum^infty_{n=1} mu({x in X : f(x) ge n}) < infty$.
My attempt at a solution: Let $A_n = {x in X : f(x) ge n}$. To show that $sum^infty_{n=1} mu ({x in X : f(x) ge n}) < infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x in X$ belong to at most finitely many $A_n$. Thus, the set ${x in X : f(x) = infty}$ has measure $0$. Now, this, together with the fact that $mu(X) < infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) le M$ for all $x$.
For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $sum^m_{n=1}mu(A_n)$, is bounded, but I'm not sure how to do so.
real-analysis analysis lebesgue-integral lebesgue-measure
real-analysis analysis lebesgue-integral lebesgue-measure
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
edited Aug 6 '15 at 19:47
poppy3345
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
asked Aug 6 '15 at 19:23
poppy3345poppy3345
993523
993523
$begingroup$
What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
$endgroup$
– user2566092
Aug 6 '15 at 19:45
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@user2566092 yes, sorry, totally screwed up there! it's been edited
$endgroup$
– poppy3345
Aug 6 '15 at 19:46
$begingroup$
@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
$endgroup$
– Gary.
Aug 6 '15 at 20:48
add a comment |
$begingroup$
What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
$endgroup$
– user2566092
Aug 6 '15 at 19:45
$begingroup$
@user2566092 yes, sorry, totally screwed up there! it's been edited
$endgroup$
– poppy3345
Aug 6 '15 at 19:46
$begingroup$
@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
$endgroup$
– Gary.
Aug 6 '15 at 20:48
$begingroup$
What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
$endgroup$
– user2566092
Aug 6 '15 at 19:45
$begingroup$
What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
$endgroup$
– user2566092
Aug 6 '15 at 19:45
$begingroup$
@user2566092 yes, sorry, totally screwed up there! it's been edited
$endgroup$
– poppy3345
Aug 6 '15 at 19:46
$begingroup$
@user2566092 yes, sorry, totally screwed up there! it's been edited
$endgroup$
– poppy3345
Aug 6 '15 at 19:46
$begingroup$
@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
$endgroup$
– Gary.
Aug 6 '15 at 20:48
$begingroup$
@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
$endgroup$
– Gary.
Aug 6 '15 at 20:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Convince yourself that $$f(x)-1leq sum_{n=1}^infty {bf 1}_{(fgeq n)}(x)leq f(x)$$
for all $xin X$, then integrate with respect to $mu$.
$endgroup$
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
add a comment |
$begingroup$
Define $A_k={x:f(x)geq k}$ (as you had done so) and $B_k={x:f(x)in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have $displaystyle X=bigcup_{k=0}^infty B_k$. Also note that $displaystyle A_n=bigcup_{k=n}^infty B_k$. This gives us $displaystyle mu(X)=sum_{k=0}^infty mu(B_k)$ and $displaystyle mu(A_n)=sum_{k=n}^infty mu(B_k)$.
Assume non-negative $f:Xrightarrow Bbb R$ is integrable, then
$$infty>int_X f dmu geq sum_{k=1}^infty kmu(B_k)= sum_{k=1}^infty mu(A_k). $$
Writing out $mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.
For the other direction, assume $displaystyle sum_{k=1}^infty mu(A_k)<infty$.
Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have
$$
begin{aligned}
int_{X} f dmu&=lim_{Nrightarrowinfty}sum_{k=0}^N int_{B_k} f dmu\
&leq lim_{Nrightarrowinfty}sum_{k=0}^N (k+1)mu(B_k) \
&=mu(X)+mu(A_1)+mu(A_2)+cdots<infty
end{aligned}
$$
Again, writing out $mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
$endgroup$
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
add a comment |
$begingroup$
If $fge 0$, you have
$$int_Omega f, dmu = int_0^infty |[f geq x]| , dx.$$
The function $xmapsto |[f geq x]|$ is decreases to $0$ at $infty$.
Can you see the rest?
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
$begingroup$
Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
$endgroup$
– poppy3345
Aug 6 '15 at 20:10
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Convince yourself that $$f(x)-1leq sum_{n=1}^infty {bf 1}_{(fgeq n)}(x)leq f(x)$$
for all $xin X$, then integrate with respect to $mu$.
$endgroup$
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
add a comment |
$begingroup$
Convince yourself that $$f(x)-1leq sum_{n=1}^infty {bf 1}_{(fgeq n)}(x)leq f(x)$$
for all $xin X$, then integrate with respect to $mu$.
$endgroup$
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
add a comment |
$begingroup$
Convince yourself that $$f(x)-1leq sum_{n=1}^infty {bf 1}_{(fgeq n)}(x)leq f(x)$$
for all $xin X$, then integrate with respect to $mu$.
$endgroup$
Convince yourself that $$f(x)-1leq sum_{n=1}^infty {bf 1}_{(fgeq n)}(x)leq f(x)$$
for all $xin X$, then integrate with respect to $mu$.
answered Aug 6 '15 at 20:22
user940
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
add a comment |
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
1
1
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
$begingroup$
Very nice solution. It also shows that in one direction we don't need $mu$ to be a finite measure.
$endgroup$
– Ramiro
Aug 6 '15 at 22:25
add a comment |
$begingroup$
Define $A_k={x:f(x)geq k}$ (as you had done so) and $B_k={x:f(x)in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have $displaystyle X=bigcup_{k=0}^infty B_k$. Also note that $displaystyle A_n=bigcup_{k=n}^infty B_k$. This gives us $displaystyle mu(X)=sum_{k=0}^infty mu(B_k)$ and $displaystyle mu(A_n)=sum_{k=n}^infty mu(B_k)$.
Assume non-negative $f:Xrightarrow Bbb R$ is integrable, then
$$infty>int_X f dmu geq sum_{k=1}^infty kmu(B_k)= sum_{k=1}^infty mu(A_k). $$
Writing out $mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.
For the other direction, assume $displaystyle sum_{k=1}^infty mu(A_k)<infty$.
Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have
$$
begin{aligned}
int_{X} f dmu&=lim_{Nrightarrowinfty}sum_{k=0}^N int_{B_k} f dmu\
&leq lim_{Nrightarrowinfty}sum_{k=0}^N (k+1)mu(B_k) \
&=mu(X)+mu(A_1)+mu(A_2)+cdots<infty
end{aligned}
$$
Again, writing out $mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
$endgroup$
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
add a comment |
$begingroup$
Define $A_k={x:f(x)geq k}$ (as you had done so) and $B_k={x:f(x)in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have $displaystyle X=bigcup_{k=0}^infty B_k$. Also note that $displaystyle A_n=bigcup_{k=n}^infty B_k$. This gives us $displaystyle mu(X)=sum_{k=0}^infty mu(B_k)$ and $displaystyle mu(A_n)=sum_{k=n}^infty mu(B_k)$.
Assume non-negative $f:Xrightarrow Bbb R$ is integrable, then
$$infty>int_X f dmu geq sum_{k=1}^infty kmu(B_k)= sum_{k=1}^infty mu(A_k). $$
Writing out $mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.
For the other direction, assume $displaystyle sum_{k=1}^infty mu(A_k)<infty$.
Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have
$$
begin{aligned}
int_{X} f dmu&=lim_{Nrightarrowinfty}sum_{k=0}^N int_{B_k} f dmu\
&leq lim_{Nrightarrowinfty}sum_{k=0}^N (k+1)mu(B_k) \
&=mu(X)+mu(A_1)+mu(A_2)+cdots<infty
end{aligned}
$$
Again, writing out $mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
$endgroup$
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
add a comment |
$begingroup$
Define $A_k={x:f(x)geq k}$ (as you had done so) and $B_k={x:f(x)in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have $displaystyle X=bigcup_{k=0}^infty B_k$. Also note that $displaystyle A_n=bigcup_{k=n}^infty B_k$. This gives us $displaystyle mu(X)=sum_{k=0}^infty mu(B_k)$ and $displaystyle mu(A_n)=sum_{k=n}^infty mu(B_k)$.
Assume non-negative $f:Xrightarrow Bbb R$ is integrable, then
$$infty>int_X f dmu geq sum_{k=1}^infty kmu(B_k)= sum_{k=1}^infty mu(A_k). $$
Writing out $mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.
For the other direction, assume $displaystyle sum_{k=1}^infty mu(A_k)<infty$.
Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have
$$
begin{aligned}
int_{X} f dmu&=lim_{Nrightarrowinfty}sum_{k=0}^N int_{B_k} f dmu\
&leq lim_{Nrightarrowinfty}sum_{k=0}^N (k+1)mu(B_k) \
&=mu(X)+mu(A_1)+mu(A_2)+cdots<infty
end{aligned}
$$
Again, writing out $mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
$endgroup$
Define $A_k={x:f(x)geq k}$ (as you had done so) and $B_k={x:f(x)in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have $displaystyle X=bigcup_{k=0}^infty B_k$. Also note that $displaystyle A_n=bigcup_{k=n}^infty B_k$. This gives us $displaystyle mu(X)=sum_{k=0}^infty mu(B_k)$ and $displaystyle mu(A_n)=sum_{k=n}^infty mu(B_k)$.
Assume non-negative $f:Xrightarrow Bbb R$ is integrable, then
$$infty>int_X f dmu geq sum_{k=1}^infty kmu(B_k)= sum_{k=1}^infty mu(A_k). $$
Writing out $mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.
For the other direction, assume $displaystyle sum_{k=1}^infty mu(A_k)<infty$.
Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have
$$
begin{aligned}
int_{X} f dmu&=lim_{Nrightarrowinfty}sum_{k=0}^N int_{B_k} f dmu\
&leq lim_{Nrightarrowinfty}sum_{k=0}^N (k+1)mu(B_k) \
&=mu(X)+mu(A_1)+mu(A_2)+cdots<infty
end{aligned}
$$
Again, writing out $mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
edited Jan 24 at 0:25
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
answered Aug 6 '15 at 21:53
jdodsjdods
3,66011234
3,66011234
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
add a comment |
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
$begingroup$
Why do you use $limlimits_{n to infty} sumlimits_{k = 0}^{n}$ instead of $sumlimits_{k = 0}^{infty}$ in the other direction?
$endgroup$
– Viktor Glombik
Jan 23 at 19:22
add a comment |
$begingroup$
If $fge 0$, you have
$$int_Omega f, dmu = int_0^infty |[f geq x]| , dx.$$
The function $xmapsto |[f geq x]|$ is decreases to $0$ at $infty$.
Can you see the rest?
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
$begingroup$
Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
$endgroup$
– poppy3345
Aug 6 '15 at 20:10
add a comment |
$begingroup$
If $fge 0$, you have
$$int_Omega f, dmu = int_0^infty |[f geq x]| , dx.$$
The function $xmapsto |[f geq x]|$ is decreases to $0$ at $infty$.
Can you see the rest?
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
$begingroup$
Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
$endgroup$
– poppy3345
Aug 6 '15 at 20:10
add a comment |
$begingroup$
If $fge 0$, you have
$$int_Omega f, dmu = int_0^infty |[f geq x]| , dx.$$
The function $xmapsto |[f geq x]|$ is decreases to $0$ at $infty$.
Can you see the rest?
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
$endgroup$
If $fge 0$, you have
$$int_Omega f, dmu = int_0^infty |[f geq x]| , dx.$$
The function $xmapsto |[f geq x]|$ is decreases to $0$ at $infty$.
Can you see the rest?
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
edited Aug 6 '15 at 20:48
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
answered Aug 6 '15 at 19:28
ncmathsadistncmathsadist
42.9k260103
42.9k260103
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I use $|E|$ for the Lebesgue measure of $E$.
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– ncmathsadist
Aug 6 '15 at 19:28
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Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
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– poppy3345
Aug 6 '15 at 20:10
add a comment |
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I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
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Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
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– poppy3345
Aug 6 '15 at 20:10
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I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
$begingroup$
I use $|E|$ for the Lebesgue measure of $E$.
$endgroup$
– ncmathsadist
Aug 6 '15 at 19:28
$begingroup$
Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
$endgroup$
– poppy3345
Aug 6 '15 at 20:10
$begingroup$
Is this saying that $int f = int m({f(x) : f(x) ge x})$? @ncmathsadist
$endgroup$
– poppy3345
Aug 6 '15 at 20:10
add a comment |
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What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation?
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– user2566092
Aug 6 '15 at 19:45
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@user2566092 yes, sorry, totally screwed up there! it's been edited
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– poppy3345
Aug 6 '15 at 19:46
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@gesa: Doesn't a convergent sum like , $Sigma_{n=1}^{infty} mu(A_n) $ have bounded partial sums?
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– Gary.
Aug 6 '15 at 20:48