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







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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

















up vote
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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







share|cite|improve this question






















  • $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















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







share|cite|improve this question













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






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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




















  • $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












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 $iin{1,dots,n}$.






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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, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      active

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      up vote
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      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 $iin{1,dots,n}$.






      share|cite|improve this answer

























        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 $iin{1,dots,n}$.






        share|cite|improve this answer























          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 $iin{1,dots,n}$.






          share|cite|improve this answer












          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 $iin{1,dots,n}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 15:42









          drhab

          96k543126




          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, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.






              share|cite|improve this answer

























                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, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.






                share|cite|improve this answer























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                  up vote
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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, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.






                  share|cite|improve this answer












                  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, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 15:20









                  John_Wick

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                  1,134111






























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