$KsubseteqBbb R^n$ compact, $f:Bbb R^nrightarrowBbb R$ Lebesgue measurable then $lim_{Rrightarrow...
$begingroup$
(Verification)As the title says, if $K subseteq Bbb R^n$ is compact and $f:Bbb R^n rightarrow Bbb R$ is a Lebesgue measurable function I would like to prove
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
I don't even know if this is true so I would like to know if my proof is correct. Given $epsilon >0$ by Lusin's Theorem there is $Fsubseteq K$ closed (therefore compact) such that $f$ is continuous over $F$ and $|K-F|< epsilon$. Then we have an $M in Bbb R_{>0}$ with $|f(x)|<M$ for every $x in F$ (because $f$ is continuous over $F$ wich is compact). Let $R>M$ then we have
$${x in K : |f(x)|>R }={x in F : |f(x)|>R } , cup , {x in K-F : |f(x)|>R }$$
It's clear ${x in F : |f(x)|>R }= emptyset$ because $|f(x)|<M<R$ for every $x in F$. Then
$$|{x in K : |f(x)|>R }|=|{x in K-F : |f(x)|>R }| le |K-F| lt epsilon$$
So for every $epsilon >0$ we found $M>0$ with $|{x in K : |f(x)|>R }|<epsilon$ for every $R ge M$. We proved
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
Is there any mistake in my proof?
real-analysis analysis lebesgue-measure
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
$endgroup$
add a comment |
$begingroup$
(Verification)As the title says, if $K subseteq Bbb R^n$ is compact and $f:Bbb R^n rightarrow Bbb R$ is a Lebesgue measurable function I would like to prove
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
I don't even know if this is true so I would like to know if my proof is correct. Given $epsilon >0$ by Lusin's Theorem there is $Fsubseteq K$ closed (therefore compact) such that $f$ is continuous over $F$ and $|K-F|< epsilon$. Then we have an $M in Bbb R_{>0}$ with $|f(x)|<M$ for every $x in F$ (because $f$ is continuous over $F$ wich is compact). Let $R>M$ then we have
$${x in K : |f(x)|>R }={x in F : |f(x)|>R } , cup , {x in K-F : |f(x)|>R }$$
It's clear ${x in F : |f(x)|>R }= emptyset$ because $|f(x)|<M<R$ for every $x in F$. Then
$$|{x in K : |f(x)|>R }|=|{x in K-F : |f(x)|>R }| le |K-F| lt epsilon$$
So for every $epsilon >0$ we found $M>0$ with $|{x in K : |f(x)|>R }|<epsilon$ for every $R ge M$. We proved
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
Is there any mistake in my proof?
real-analysis analysis lebesgue-measure
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
$endgroup$
add a comment |
$begingroup$
(Verification)As the title says, if $K subseteq Bbb R^n$ is compact and $f:Bbb R^n rightarrow Bbb R$ is a Lebesgue measurable function I would like to prove
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
I don't even know if this is true so I would like to know if my proof is correct. Given $epsilon >0$ by Lusin's Theorem there is $Fsubseteq K$ closed (therefore compact) such that $f$ is continuous over $F$ and $|K-F|< epsilon$. Then we have an $M in Bbb R_{>0}$ with $|f(x)|<M$ for every $x in F$ (because $f$ is continuous over $F$ wich is compact). Let $R>M$ then we have
$${x in K : |f(x)|>R }={x in F : |f(x)|>R } , cup , {x in K-F : |f(x)|>R }$$
It's clear ${x in F : |f(x)|>R }= emptyset$ because $|f(x)|<M<R$ for every $x in F$. Then
$$|{x in K : |f(x)|>R }|=|{x in K-F : |f(x)|>R }| le |K-F| lt epsilon$$
So for every $epsilon >0$ we found $M>0$ with $|{x in K : |f(x)|>R }|<epsilon$ for every $R ge M$. We proved
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
Is there any mistake in my proof?
real-analysis analysis lebesgue-measure
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
$endgroup$
(Verification)As the title says, if $K subseteq Bbb R^n$ is compact and $f:Bbb R^n rightarrow Bbb R$ is a Lebesgue measurable function I would like to prove
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
I don't even know if this is true so I would like to know if my proof is correct. Given $epsilon >0$ by Lusin's Theorem there is $Fsubseteq K$ closed (therefore compact) such that $f$ is continuous over $F$ and $|K-F|< epsilon$. Then we have an $M in Bbb R_{>0}$ with $|f(x)|<M$ for every $x in F$ (because $f$ is continuous over $F$ wich is compact). Let $R>M$ then we have
$${x in K : |f(x)|>R }={x in F : |f(x)|>R } , cup , {x in K-F : |f(x)|>R }$$
It's clear ${x in F : |f(x)|>R }= emptyset$ because $|f(x)|<M<R$ for every $x in F$. Then
$$|{x in K : |f(x)|>R }|=|{x in K-F : |f(x)|>R }| le |K-F| lt epsilon$$
So for every $epsilon >0$ we found $M>0$ with $|{x in K : |f(x)|>R }|<epsilon$ for every $R ge M$. We proved
$$lim_{R rightarrow +infty} |{ x in K : |f(x)|>R }|=0$$
Is there any mistake in my proof?
real-analysis analysis lebesgue-measure
real-analysis analysis lebesgue-measure
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
asked Dec 16 '18 at 20:54
Marcos Martínez WagnerMarcos Martínez Wagner
1148
1148
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