If $fin L^1(mu)$ then $f$ is finite $mu$-a.e.
$begingroup$
Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).
My proof:
Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$
I intend to show $mu(A)=0$
As $fin L^{1}(mu)$
$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$
Applying Chebyshev inequality:
$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $
Questions:
Is my proof right? If not why not?
What are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
edited Jan 11 at 19:54
6005
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
$endgroup$
add a comment |
$begingroup$
Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).
My proof:
Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$
I intend to show $mu(A)=0$
As $fin L^{1}(mu)$
$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$
Applying Chebyshev inequality:
$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $
Questions:
Is my proof right? If not why not?
What are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
edited Jan 11 at 19:54
6005
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
$endgroup$
$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39
add a comment |
$begingroup$
Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).
My proof:
Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$
I intend to show $mu(A)=0$
As $fin L^{1}(mu)$
$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$
Applying Chebyshev inequality:
$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $
Questions:
Is my proof right? If not why not?
What are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
edited Jan 11 at 19:54
6005
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
$endgroup$
Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).
My proof:
Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$
I intend to show $mu(A)=0$
As $fin L^{1}(mu)$
$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$
Applying Chebyshev inequality:
$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $
Questions:
Is my proof right? If not why not?
What are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
measure-theory proof-verification proof-writing
edited Jan 11 at 19:54
6005
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
edited Jan 11 at 19:54
6005
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
edited Jan 11 at 19:54
6005
37.1k752127
edited Jan 11 at 19:54
6005
37.1k752127
edited Jan 11 at 19:54
6005
37.1k752127
37.1k752127
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
asked Jan 11 at 19:43
Pedro GomesPedro Gomes
2,0062821
2,0062821
$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39
add a comment |
$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39
$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nice proof—it looks right to me.
What are the alternatives?
Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nice proof—it looks right to me.
What are the alternatives?
Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
$endgroup$
add a comment |
$begingroup$
Nice proof—it looks right to me.
What are the alternatives?
Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
$endgroup$
add a comment |
$begingroup$
Nice proof—it looks right to me.
What are the alternatives?
Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
$endgroup$
Nice proof—it looks right to me.
What are the alternatives?
Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
edited Jan 11 at 20:31
Alex Ortiz
11.4k21442
11.4k21442
answered Jan 11 at 19:50
60056005
37.1k752127
answered Jan 11 at 19:50
60056005
37.1k752127
answered Jan 11 at 19:50
60056005
37.1k752127
37.1k752127
add a comment |
add a comment |
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$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45
$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09
$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38
$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39