Lebesgue measure: $limlimits_{n to infty} n cdot m(S_n)=0$ where $S_n={x in E mid |f(x)|geq n} $
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Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$
We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.
measure-theory lebesgue-integral lebesgue-measure
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
add a comment |
up vote
2
down vote
favorite
Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$
We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.
measure-theory lebesgue-integral lebesgue-measure
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$
We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.
measure-theory lebesgue-integral lebesgue-measure
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$
We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:25
Martin Sleziak
44.6k7115269
44.6k7115269
asked Nov 27 at 6:04
ICanMakeYouHateME
154
asked Nov 27 at 6:04
ICanMakeYouHateME
154
asked Nov 27 at 6:04
ICanMakeYouHateME
154
154
Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27
add a comment |
Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27
Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27
Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.
answered Nov 27 at 6:16
Bungo
13.6k22148
add a comment |
up vote
3
down vote
Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.
answered Nov 27 at 6:12
Dante Grevino
90117
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.
answered Nov 27 at 6:16
Bungo
13.6k22148
add a comment |
up vote
3
down vote
accepted
Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.
answered Nov 27 at 6:16
Bungo
13.6k22148
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.
answered Nov 27 at 6:16
Bungo
13.6k22148
Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.
answered Nov 27 at 6:16
Bungo
13.6k22148
answered Nov 27 at 6:16
Bungo
13.6k22148
answered Nov 27 at 6:16
Bungo
13.6k22148
answered Nov 27 at 6:16
Bungo
13.6k22148
13.6k22148
add a comment |
add a comment |
up vote
3
down vote
Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.
answered Nov 27 at 6:12
Dante Grevino
90117
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
add a comment |
up vote
3
down vote
Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.
answered Nov 27 at 6:12
Dante Grevino
90117
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.
answered Nov 27 at 6:12
Dante Grevino
90117
Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.
answered Nov 27 at 6:12
Dante Grevino
90117
answered Nov 27 at 6:12
Dante Grevino
90117
answered Nov 27 at 6:12
Dante Grevino
90117
answered Nov 27 at 6:12
Dante Grevino
90117
90117
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
add a comment |
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
– ICanMakeYouHateME
Nov 27 at 6:18
1
1
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
– Dante Grevino
Nov 27 at 6:21
add a comment |
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Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27