Why is the set $[0le f_1le f_2]$ measurable?
$begingroup$
This question comes reading Analysis III of Amann and Escher. We set
$$[flealpha]:={xin X: f(x)lealpha}$$
Then it is stated that if $f_1,f_2inmathcal L_0(X,lambda_n,overline{Bbb R}^+)$, that is, each $f_k$ is Lebesgue measurable for some $XsubsetBbb R^n$ Lebesgue measurable, then the set
$$[0le f_1le f_2]:={xin X: 0le f_1(x)le f_2(x)}tag1$$
is $lambda_n$-measurable. The book says that this is implied by the fact that
$$finmathcal L_0(X,lambda_m,overline{Bbb R})iff forallalphainoverline{Bbb R}:[flealpha]inmathcal L(m)tag2$$
where $mathcal L(m)$ is the Lebesgue $sigma$-algebra in $Bbb R^m$. However I dont see how $(2)$ imply $(1)$. can someone show me the relation? Or, if this is not clear, can someone show me other way to see why $(1)$ is measurable?
real-analysis measure-theory lebesgue-measure
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
This question comes reading Analysis III of Amann and Escher. We set
$$[flealpha]:={xin X: f(x)lealpha}$$
Then it is stated that if $f_1,f_2inmathcal L_0(X,lambda_n,overline{Bbb R}^+)$, that is, each $f_k$ is Lebesgue measurable for some $XsubsetBbb R^n$ Lebesgue measurable, then the set
$$[0le f_1le f_2]:={xin X: 0le f_1(x)le f_2(x)}tag1$$
is $lambda_n$-measurable. The book says that this is implied by the fact that
$$finmathcal L_0(X,lambda_m,overline{Bbb R})iff forallalphainoverline{Bbb R}:[flealpha]inmathcal L(m)tag2$$
where $mathcal L(m)$ is the Lebesgue $sigma$-algebra in $Bbb R^m$. However I dont see how $(2)$ imply $(1)$. can someone show me the relation? Or, if this is not clear, can someone show me other way to see why $(1)$ is measurable?
real-analysis measure-theory lebesgue-measure
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
This question comes reading Analysis III of Amann and Escher. We set
$$[flealpha]:={xin X: f(x)lealpha}$$
Then it is stated that if $f_1,f_2inmathcal L_0(X,lambda_n,overline{Bbb R}^+)$, that is, each $f_k$ is Lebesgue measurable for some $XsubsetBbb R^n$ Lebesgue measurable, then the set
$$[0le f_1le f_2]:={xin X: 0le f_1(x)le f_2(x)}tag1$$
is $lambda_n$-measurable. The book says that this is implied by the fact that
$$finmathcal L_0(X,lambda_m,overline{Bbb R})iff forallalphainoverline{Bbb R}:[flealpha]inmathcal L(m)tag2$$
where $mathcal L(m)$ is the Lebesgue $sigma$-algebra in $Bbb R^m$. However I dont see how $(2)$ imply $(1)$. can someone show me the relation? Or, if this is not clear, can someone show me other way to see why $(1)$ is measurable?
real-analysis measure-theory lebesgue-measure
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
$endgroup$
This question comes reading Analysis III of Amann and Escher. We set
$$[flealpha]:={xin X: f(x)lealpha}$$
Then it is stated that if $f_1,f_2inmathcal L_0(X,lambda_n,overline{Bbb R}^+)$, that is, each $f_k$ is Lebesgue measurable for some $XsubsetBbb R^n$ Lebesgue measurable, then the set
$$[0le f_1le f_2]:={xin X: 0le f_1(x)le f_2(x)}tag1$$
is $lambda_n$-measurable. The book says that this is implied by the fact that
$$finmathcal L_0(X,lambda_m,overline{Bbb R})iff forallalphainoverline{Bbb R}:[flealpha]inmathcal L(m)tag2$$
where $mathcal L(m)$ is the Lebesgue $sigma$-algebra in $Bbb R^m$. However I dont see how $(2)$ imply $(1)$. can someone show me the relation? Or, if this is not clear, can someone show me other way to see why $(1)$ is measurable?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
edited Dec 12 '18 at 14:53
Masacroso
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
asked Dec 12 '18 at 14:17
MasacrosoMasacroso
13.1k41747
13.1k41747
add a comment |
add a comment |
1 Answer
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$begingroup$
It is the intersection of the sets
$$ [f_1 ge 0], [g ge 0]$$
Where $g$ is the measurable function $g=f_2-f_1$.
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is the intersection of the sets
$$ [f_1 ge 0], [g ge 0]$$
Where $g$ is the measurable function $g=f_2-f_1$.
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
add a comment |
$begingroup$
It is the intersection of the sets
$$ [f_1 ge 0], [g ge 0]$$
Where $g$ is the measurable function $g=f_2-f_1$.
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
add a comment |
$begingroup$
It is the intersection of the sets
$$ [f_1 ge 0], [g ge 0]$$
Where $g$ is the measurable function $g=f_2-f_1$.
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
It is the intersection of the sets
$$ [f_1 ge 0], [g ge 0]$$
Where $g$ is the measurable function $g=f_2-f_1$.
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 14:58
Calvin KhorCalvin Khor
12.1k21438
12.1k21438
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
add a comment |
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
$begingroup$
oh, my god... what a dumb... yes, we can write $0le f_1le f_2$ as $0le f_1,land, f_1le f_2$
$endgroup$
– Masacroso
Dec 12 '18 at 15:01
1
1
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
$begingroup$
@Masacroso yes, and yours is better actually since you allow the functions to take the value + infinity
$endgroup$
– Calvin Khor
Dec 12 '18 at 15:04
add a comment |
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