Prove that region under graph of function is measurable












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In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).



The setting: $(X,mathcal{A}, mu)$ is a $sigma$-finite measure space, $lambda$ is Lebesgue measure on $(mathbb{R},mathcal{B}(mathbb{R}))$ (Borel $sigma$-algebra), $f:X to [0,+infty]$ is $mathcal{A}$-measurable, and we are considering the region under the graph of $f$,



$E={(x,y)in X times mathbb{R}|0leq y < f(x)}$.



I need to prove $E in mathcal{A} times mathcal{B}(mathbb{R})$. I thought to write $E=g^{-1}((0,+infty])cap(X times [0,+infty])$ where $g(x,y)=f(x)-y$ but I can't see why $g$ must be $mathcal{A} times mathcal{B}(mathbb{R})$-measurable. Any help would be appreciated.






analysis measure-theory graphing-functions







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    2












    $begingroup$


    In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).



    The setting: $(X,mathcal{A}, mu)$ is a $sigma$-finite measure space, $lambda$ is Lebesgue measure on $(mathbb{R},mathcal{B}(mathbb{R}))$ (Borel $sigma$-algebra), $f:X to [0,+infty]$ is $mathcal{A}$-measurable, and we are considering the region under the graph of $f$,



    $E={(x,y)in X times mathbb{R}|0leq y < f(x)}$.



    I need to prove $E in mathcal{A} times mathcal{B}(mathbb{R})$. I thought to write $E=g^{-1}((0,+infty])cap(X times [0,+infty])$ where $g(x,y)=f(x)-y$ but I can't see why $g$ must be $mathcal{A} times mathcal{B}(mathbb{R})$-measurable. Any help would be appreciated.






    analysis measure-theory graphing-functions







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).



      The setting: $(X,mathcal{A}, mu)$ is a $sigma$-finite measure space, $lambda$ is Lebesgue measure on $(mathbb{R},mathcal{B}(mathbb{R}))$ (Borel $sigma$-algebra), $f:X to [0,+infty]$ is $mathcal{A}$-measurable, and we are considering the region under the graph of $f$,



      $E={(x,y)in X times mathbb{R}|0leq y < f(x)}$.



      I need to prove $E in mathcal{A} times mathcal{B}(mathbb{R})$. I thought to write $E=g^{-1}((0,+infty])cap(X times [0,+infty])$ where $g(x,y)=f(x)-y$ but I can't see why $g$ must be $mathcal{A} times mathcal{B}(mathbb{R})$-measurable. Any help would be appreciated.






      analysis measure-theory graphing-functions







      share|cite|improve this question









      $endgroup$




      In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).



      The setting: $(X,mathcal{A}, mu)$ is a $sigma$-finite measure space, $lambda$ is Lebesgue measure on $(mathbb{R},mathcal{B}(mathbb{R}))$ (Borel $sigma$-algebra), $f:X to [0,+infty]$ is $mathcal{A}$-measurable, and we are considering the region under the graph of $f$,



      $E={(x,y)in X times mathbb{R}|0leq y < f(x)}$.



      I need to prove $E in mathcal{A} times mathcal{B}(mathbb{R})$. I thought to write $E=g^{-1}((0,+infty])cap(X times [0,+infty])$ where $g(x,y)=f(x)-y$ but I can't see why $g$ must be $mathcal{A} times mathcal{B}(mathbb{R})$-measurable. Any help would be appreciated.






      analysis measure-theory graphing-functions




      analysis measure-theory graphing-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 6 '18 at 21:52









      AlephNullAlephNull

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          $begingroup$

          $g=kcirc h$ where $h(x,y)=(f(x),y)$ and $k(a,b)=a-b$. [ Here $h:Xtimes mathbb R to mathbb R^{2}$ and $k:mathbb R^{2} to mathbb R$]. $k:mathbb R^{2} to mathbb R$ is Borel measurable because it is continuous. To show that $h$ is measurable it is enough to show that $h^{-1} (A times B) in mathcal A times B(mathbb R)$ for $A,B in mathcal B(mathbb R)$. This is clear because $h^{-1} (A times B)=f^{-1}(A) times B$.



          I have assumed that $f$ takes only finite values. To handle the general case let $g(x)=f(x)$ if $f(x) <infty$ and $0$ if $f(x)=infty$. Let $F={(x,y):0leq y <g(x)}$. Then $E=(f^{-1}{infty}times [0,infty)) cup [(f^{-1}{mathbb R}times mathbb R) cap F]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 0:36










          • $begingroup$
            @AlephNull I have edited my answer.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 5:19










          • $begingroup$
            Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 11:25











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          active

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          $begingroup$

          $g=kcirc h$ where $h(x,y)=(f(x),y)$ and $k(a,b)=a-b$. [ Here $h:Xtimes mathbb R to mathbb R^{2}$ and $k:mathbb R^{2} to mathbb R$]. $k:mathbb R^{2} to mathbb R$ is Borel measurable because it is continuous. To show that $h$ is measurable it is enough to show that $h^{-1} (A times B) in mathcal A times B(mathbb R)$ for $A,B in mathcal B(mathbb R)$. This is clear because $h^{-1} (A times B)=f^{-1}(A) times B$.



          I have assumed that $f$ takes only finite values. To handle the general case let $g(x)=f(x)$ if $f(x) <infty$ and $0$ if $f(x)=infty$. Let $F={(x,y):0leq y <g(x)}$. Then $E=(f^{-1}{infty}times [0,infty)) cup [(f^{-1}{mathbb R}times mathbb R) cap F]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 0:36










          • $begingroup$
            @AlephNull I have edited my answer.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 5:19










          • $begingroup$
            Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 11:25
















          1












          $begingroup$

          $g=kcirc h$ where $h(x,y)=(f(x),y)$ and $k(a,b)=a-b$. [ Here $h:Xtimes mathbb R to mathbb R^{2}$ and $k:mathbb R^{2} to mathbb R$]. $k:mathbb R^{2} to mathbb R$ is Borel measurable because it is continuous. To show that $h$ is measurable it is enough to show that $h^{-1} (A times B) in mathcal A times B(mathbb R)$ for $A,B in mathcal B(mathbb R)$. This is clear because $h^{-1} (A times B)=f^{-1}(A) times B$.



          I have assumed that $f$ takes only finite values. To handle the general case let $g(x)=f(x)$ if $f(x) <infty$ and $0$ if $f(x)=infty$. Let $F={(x,y):0leq y <g(x)}$. Then $E=(f^{-1}{infty}times [0,infty)) cup [(f^{-1}{mathbb R}times mathbb R) cap F]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 0:36










          • $begingroup$
            @AlephNull I have edited my answer.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 5:19










          • $begingroup$
            Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 11:25














          1












          1








          1





          $begingroup$

          $g=kcirc h$ where $h(x,y)=(f(x),y)$ and $k(a,b)=a-b$. [ Here $h:Xtimes mathbb R to mathbb R^{2}$ and $k:mathbb R^{2} to mathbb R$]. $k:mathbb R^{2} to mathbb R$ is Borel measurable because it is continuous. To show that $h$ is measurable it is enough to show that $h^{-1} (A times B) in mathcal A times B(mathbb R)$ for $A,B in mathcal B(mathbb R)$. This is clear because $h^{-1} (A times B)=f^{-1}(A) times B$.



          I have assumed that $f$ takes only finite values. To handle the general case let $g(x)=f(x)$ if $f(x) <infty$ and $0$ if $f(x)=infty$. Let $F={(x,y):0leq y <g(x)}$. Then $E=(f^{-1}{infty}times [0,infty)) cup [(f^{-1}{mathbb R}times mathbb R) cap F]$.






          share|cite|improve this answer











          $endgroup$



          $g=kcirc h$ where $h(x,y)=(f(x),y)$ and $k(a,b)=a-b$. [ Here $h:Xtimes mathbb R to mathbb R^{2}$ and $k:mathbb R^{2} to mathbb R$]. $k:mathbb R^{2} to mathbb R$ is Borel measurable because it is continuous. To show that $h$ is measurable it is enough to show that $h^{-1} (A times B) in mathcal A times B(mathbb R)$ for $A,B in mathcal B(mathbb R)$. This is clear because $h^{-1} (A times B)=f^{-1}(A) times B$.



          I have assumed that $f$ takes only finite values. To handle the general case let $g(x)=f(x)$ if $f(x) <infty$ and $0$ if $f(x)=infty$. Let $F={(x,y):0leq y <g(x)}$. Then $E=(f^{-1}{infty}times [0,infty)) cup [(f^{-1}{mathbb R}times mathbb R) cap F]$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 5:19

























          answered Dec 6 '18 at 23:46









          Kavi Rama MurthyKavi Rama Murthy

          59k42161




          59k42161












          • $begingroup$
            I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 0:36










          • $begingroup$
            @AlephNull I have edited my answer.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 5:19










          • $begingroup$
            Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 11:25


















          • $begingroup$
            I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 0:36










          • $begingroup$
            @AlephNull I have edited my answer.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 5:19










          • $begingroup$
            Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
            $endgroup$
            – AlephNull
            Dec 7 '18 at 11:25
















          $begingroup$
          I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
          $endgroup$
          – AlephNull
          Dec 7 '18 at 0:36




          $begingroup$
          I suppose you mean $h^{-1}(A times B) in mathcal{A} times mathcal{B}(mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case.
          $endgroup$
          – AlephNull
          Dec 7 '18 at 0:36












          $begingroup$
          @AlephNull I have edited my answer.
          $endgroup$
          – Kavi Rama Murthy
          Dec 7 '18 at 5:19




          $begingroup$
          @AlephNull I have edited my answer.
          $endgroup$
          – Kavi Rama Murthy
          Dec 7 '18 at 5:19












          $begingroup$
          Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
          $endgroup$
          – AlephNull
          Dec 7 '18 at 11:25




          $begingroup$
          Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one.
          $endgroup$
          – AlephNull
          Dec 7 '18 at 11:25


















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