Lebesgue integral (measure zero)
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Let $(X, mathcal{A},mu)$ be a measure space and $h:X to mathbb{R}, h geq0$ a measurable function.
Define a map $mu_h:mathcal{A} to mathbb{R}, mu_h(E):= int_E h dmu$ for $E in mathcal{A}$.
How can it be shown that if $mu(E)=0 Rightarrow mu_h(E)=0$?
To prove this implication I tried:
$int_E h dmu=int_E h_+ dmu-int_E h_- dmu$
So $int_E h_+ dmu=int h_+1_E dmu=$sup$int e dmu$, with $e$ is a simple function.
Here I don't know how to continue to show this implication. Or is there another way to prove it?
measure-theory lebesgue-integral
asked Nov 22 at 15:01
Olsgur
444
add a comment |
up vote
0
down vote
favorite
Let $(X, mathcal{A},mu)$ be a measure space and $h:X to mathbb{R}, h geq0$ a measurable function.
Define a map $mu_h:mathcal{A} to mathbb{R}, mu_h(E):= int_E h dmu$ for $E in mathcal{A}$.
How can it be shown that if $mu(E)=0 Rightarrow mu_h(E)=0$?
To prove this implication I tried:
$int_E h dmu=int_E h_+ dmu-int_E h_- dmu$
So $int_E h_+ dmu=int h_+1_E dmu=$sup$int e dmu$, with $e$ is a simple function.
Here I don't know how to continue to show this implication. Or is there another way to prove it?
measure-theory lebesgue-integral
asked Nov 22 at 15:01
Olsgur
444
$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X, mathcal{A},mu)$ be a measure space and $h:X to mathbb{R}, h geq0$ a measurable function.
Define a map $mu_h:mathcal{A} to mathbb{R}, mu_h(E):= int_E h dmu$ for $E in mathcal{A}$.
How can it be shown that if $mu(E)=0 Rightarrow mu_h(E)=0$?
To prove this implication I tried:
$int_E h dmu=int_E h_+ dmu-int_E h_- dmu$
So $int_E h_+ dmu=int h_+1_E dmu=$sup$int e dmu$, with $e$ is a simple function.
Here I don't know how to continue to show this implication. Or is there another way to prove it?
measure-theory lebesgue-integral
asked Nov 22 at 15:01
Olsgur
444
Let $(X, mathcal{A},mu)$ be a measure space and $h:X to mathbb{R}, h geq0$ a measurable function.
Define a map $mu_h:mathcal{A} to mathbb{R}, mu_h(E):= int_E h dmu$ for $E in mathcal{A}$.
How can it be shown that if $mu(E)=0 Rightarrow mu_h(E)=0$?
To prove this implication I tried:
$int_E h dmu=int_E h_+ dmu-int_E h_- dmu$
So $int_E h_+ dmu=int h_+1_E dmu=$sup$int e dmu$, with $e$ is a simple function.
Here I don't know how to continue to show this implication. Or is there another way to prove it?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 22 at 15:01
Olsgur
444
asked Nov 22 at 15:01
Olsgur
444
asked Nov 22 at 15:01
Olsgur
444
asked Nov 22 at 15:01
Olsgur
444
asked Nov 22 at 15:01
Olsgur
444
444
$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16
add a comment |
$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16
$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16
$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16
add a comment |
2 Answers
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By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$
So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.
Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.
answered Nov 22 at 15:42
drhab
96k543126
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The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.
So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.
answered Nov 22 at 15:20
John_Wick
1,134111
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$
So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.
Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.
answered Nov 22 at 15:42
drhab
96k543126
add a comment |
up vote
0
down vote
accepted
By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$
So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.
Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.
answered Nov 22 at 15:42
drhab
96k543126
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$
So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.
Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.
answered Nov 22 at 15:42
drhab
96k543126
By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$
So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.
Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.
answered Nov 22 at 15:42
drhab
96k543126
answered Nov 22 at 15:42
drhab
96k543126
answered Nov 22 at 15:42
drhab
96k543126
answered Nov 22 at 15:42
drhab
96k543126
96k543126
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add a comment |
up vote
0
down vote
The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.
So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.
answered Nov 22 at 15:20
John_Wick
1,134111
add a comment |
up vote
0
down vote
The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.
So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.
answered Nov 22 at 15:20
John_Wick
1,134111
add a comment |
up vote
0
down vote
up vote
0
down vote
The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.
So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.
answered Nov 22 at 15:20
John_Wick
1,134111
The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.
So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.
answered Nov 22 at 15:20
John_Wick
1,134111
answered Nov 22 at 15:20
John_Wick
1,134111
answered Nov 22 at 15:20
John_Wick
1,134111
answered Nov 22 at 15:20
John_Wick
1,134111
1,134111
add a comment |
add a comment |
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$int _E h dmu := int h mathbf 1_E dmu $ and $h mathbf 1_E = 0 $ $mu$-a.e.
– Calvin Khor
Nov 22 at 15:16