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If $finmathcal{L}^1(mu)$, prove that ${xin X: f(x)neq 0}$ is the countable union of sets with finite...
Tonight I am working on the following problem:
Suppose $(X,mathcal{S},mu)$ is a measure space and $finmathcal{L}^1(mu)$. Prove that $${xin X: f(x)neq 0}$$ is the countable union of sets with finite $mu$-measure.
My first reaction was to write down the definition of what it means for $f$ to be in $mathcal{L}^1(mu)$:
Suppose $(X,mathcal{S},mu)$ is a measure space. An $mathcal{S}$-measurable function $f:Xrightarrowmathbf{R}$ is contained in $mathcal{L}^1(mu)$ if $$||f||_1=int|f|,dmu<infty.$$
Okay, so $int|f|,dmu<infty$. How does this help me? I'm not sure. After alot of searching, I read on what it means for a measure $mu$ to be $sigma$-finite. That got me thinking -- does $finmathcal{L}^1(mu)$ imply that $mu$ is $sigma$-finite? If so, then I think I could easily prove this using any of the conditions outlined in that Wikipedia page.
Am I on the right track? Any suggestions?
Thanks in advance!
real-analysis measure-theory
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
add a comment |
Tonight I am working on the following problem:
Suppose $(X,mathcal{S},mu)$ is a measure space and $finmathcal{L}^1(mu)$. Prove that $${xin X: f(x)neq 0}$$ is the countable union of sets with finite $mu$-measure.
My first reaction was to write down the definition of what it means for $f$ to be in $mathcal{L}^1(mu)$:
Suppose $(X,mathcal{S},mu)$ is a measure space. An $mathcal{S}$-measurable function $f:Xrightarrowmathbf{R}$ is contained in $mathcal{L}^1(mu)$ if $$||f||_1=int|f|,dmu<infty.$$
Okay, so $int|f|,dmu<infty$. How does this help me? I'm not sure. After alot of searching, I read on what it means for a measure $mu$ to be $sigma$-finite. That got me thinking -- does $finmathcal{L}^1(mu)$ imply that $mu$ is $sigma$-finite? If so, then I think I could easily prove this using any of the conditions outlined in that Wikipedia page.
Am I on the right track? Any suggestions?
Thanks in advance!
real-analysis measure-theory
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
add a comment |
Tonight I am working on the following problem:
Suppose $(X,mathcal{S},mu)$ is a measure space and $finmathcal{L}^1(mu)$. Prove that $${xin X: f(x)neq 0}$$ is the countable union of sets with finite $mu$-measure.
My first reaction was to write down the definition of what it means for $f$ to be in $mathcal{L}^1(mu)$:
Suppose $(X,mathcal{S},mu)$ is a measure space. An $mathcal{S}$-measurable function $f:Xrightarrowmathbf{R}$ is contained in $mathcal{L}^1(mu)$ if $$||f||_1=int|f|,dmu<infty.$$
Okay, so $int|f|,dmu<infty$. How does this help me? I'm not sure. After alot of searching, I read on what it means for a measure $mu$ to be $sigma$-finite. That got me thinking -- does $finmathcal{L}^1(mu)$ imply that $mu$ is $sigma$-finite? If so, then I think I could easily prove this using any of the conditions outlined in that Wikipedia page.
Am I on the right track? Any suggestions?
Thanks in advance!
real-analysis measure-theory
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
Tonight I am working on the following problem:
Suppose $(X,mathcal{S},mu)$ is a measure space and $finmathcal{L}^1(mu)$. Prove that $${xin X: f(x)neq 0}$$ is the countable union of sets with finite $mu$-measure.
My first reaction was to write down the definition of what it means for $f$ to be in $mathcal{L}^1(mu)$:
Suppose $(X,mathcal{S},mu)$ is a measure space. An $mathcal{S}$-measurable function $f:Xrightarrowmathbf{R}$ is contained in $mathcal{L}^1(mu)$ if $$||f||_1=int|f|,dmu<infty.$$
Okay, so $int|f|,dmu<infty$. How does this help me? I'm not sure. After alot of searching, I read on what it means for a measure $mu$ to be $sigma$-finite. That got me thinking -- does $finmathcal{L}^1(mu)$ imply that $mu$ is $sigma$-finite? If so, then I think I could easily prove this using any of the conditions outlined in that Wikipedia page.
Am I on the right track? Any suggestions?
Thanks in advance!
real-analysis measure-theory
real-analysis measure-theory
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
asked Nov 20 '18 at 4:57
Thy Art is Math
484211
484211
add a comment |
add a comment |
1 Answer
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This is a standard "layer cake" argument. Define
$$E_n={xin X,|, |f(x)|ge frac{1}{n}}$$
and note that $mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
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oldest
votes
This is a standard "layer cake" argument. Define
$$E_n={xin X,|, |f(x)|ge frac{1}{n}}$$
and note that $mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
add a comment |
This is a standard "layer cake" argument. Define
$$E_n={xin X,|, |f(x)|ge frac{1}{n}}$$
and note that $mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
add a comment |
This is a standard "layer cake" argument. Define
$$E_n={xin X,|, |f(x)|ge frac{1}{n}}$$
and note that $mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
This is a standard "layer cake" argument. Define
$$E_n={xin X,|, |f(x)|ge frac{1}{n}}$$
and note that $mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
answered Nov 20 '18 at 5:12
Ashwin Trisal
1,2141516
1,2141516
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
add a comment |
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
Ah, interesting! So where do the $E_n$'s live? In $mathcal{S}$? Also, what is the intuition behind defining $E_n$ like that? Why do we care where $|f(x)|geq frac{1}{n}$?
– Thy Art is Math
Nov 20 '18 at 5:14
1
1
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
The $E_n$s are measurable subsets of $X$, so they are elements of $mathcal S$. We define the $E_n$ in this way because they have to have finite measure, and if $f(x)neq 0$, $x$ would have to be in some $E_n$ by the Archimedean principle.
– Ashwin Trisal
Nov 20 '18 at 5:52
add a comment |
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