Showing that a set is Lebesgue Measurable in Higher Dimensions and Applying Fubini's Theorem












3














I have an idea of how to proceed, but I'm suspicious that my efforts were of no use. Let $Ainmathcal{M}$ be Lebesgue measurable, and let $g,h:A rightarrow bar{mathbb{R}}$ be Lebesgue measurable functions with $g(x)leq h(x) forall x in A$. With this, we define $E={(x,y) in mathbb{R^2} : g(x) leq y leq h(x)}$. Prove that $Ein mathcal{M^2}=mathcal{M} times mathcal{M}$.



And following this, show that if $f:E rightarrow [0,infty]$ is $mathcal{M^2}$ measurable, then
$displaystyleint_Ef dm^2=int_A(int_{g(x)}^{h(x)}f(x,y) dy) dx$, where $m^2=m times m$.



For the first part, may we write $E={(x,y) in mathbb{R^2} : g(x) leq y} cap {(x,y) in mathbb{R^2} : y leq h(x)}$
$Rightarrow E=bigcuplimits_{q in mathbb{Q}} {(x,y) in mathbb{R^2} : g(x) leq q leq y} cap {(x,y) in mathbb{R^2} : y leq q leq h(x)}$



and since $h$ and $g$ are measurable, this is a countable union of measurable sets? I don't feel so good about this reasoning, as I'm pretty sure that I would have to find a way to rewrite $E$ as cartesian product of 2 Lebesgue measurable sets (ideally, one with respect to x and the other with respect to y).
So perhaps $E=(g^{-1}[-infty,y]cap h^{-1}[y,infty]) times {y}$ works instead?



As for the second part, I'm not particularly sure how to use Fubini's in this scenario. I'd be very gracious for any help and guidance.






real-analysis measure-theory lebesgue-integral lebesgue-measure measurable-functions







share|cite|improve this question





























    3














    I have an idea of how to proceed, but I'm suspicious that my efforts were of no use. Let $Ainmathcal{M}$ be Lebesgue measurable, and let $g,h:A rightarrow bar{mathbb{R}}$ be Lebesgue measurable functions with $g(x)leq h(x) forall x in A$. With this, we define $E={(x,y) in mathbb{R^2} : g(x) leq y leq h(x)}$. Prove that $Ein mathcal{M^2}=mathcal{M} times mathcal{M}$.



    And following this, show that if $f:E rightarrow [0,infty]$ is $mathcal{M^2}$ measurable, then
    $displaystyleint_Ef dm^2=int_A(int_{g(x)}^{h(x)}f(x,y) dy) dx$, where $m^2=m times m$.



    For the first part, may we write $E={(x,y) in mathbb{R^2} : g(x) leq y} cap {(x,y) in mathbb{R^2} : y leq h(x)}$
    $Rightarrow E=bigcuplimits_{q in mathbb{Q}} {(x,y) in mathbb{R^2} : g(x) leq q leq y} cap {(x,y) in mathbb{R^2} : y leq q leq h(x)}$



    and since $h$ and $g$ are measurable, this is a countable union of measurable sets? I don't feel so good about this reasoning, as I'm pretty sure that I would have to find a way to rewrite $E$ as cartesian product of 2 Lebesgue measurable sets (ideally, one with respect to x and the other with respect to y).
    So perhaps $E=(g^{-1}[-infty,y]cap h^{-1}[y,infty]) times {y}$ works instead?



    As for the second part, I'm not particularly sure how to use Fubini's in this scenario. I'd be very gracious for any help and guidance.






    real-analysis measure-theory lebesgue-integral lebesgue-measure measurable-functions







    share|cite|improve this question



























      3












      3








      3







      I have an idea of how to proceed, but I'm suspicious that my efforts were of no use. Let $Ainmathcal{M}$ be Lebesgue measurable, and let $g,h:A rightarrow bar{mathbb{R}}$ be Lebesgue measurable functions with $g(x)leq h(x) forall x in A$. With this, we define $E={(x,y) in mathbb{R^2} : g(x) leq y leq h(x)}$. Prove that $Ein mathcal{M^2}=mathcal{M} times mathcal{M}$.



      And following this, show that if $f:E rightarrow [0,infty]$ is $mathcal{M^2}$ measurable, then
      $displaystyleint_Ef dm^2=int_A(int_{g(x)}^{h(x)}f(x,y) dy) dx$, where $m^2=m times m$.



      For the first part, may we write $E={(x,y) in mathbb{R^2} : g(x) leq y} cap {(x,y) in mathbb{R^2} : y leq h(x)}$
      $Rightarrow E=bigcuplimits_{q in mathbb{Q}} {(x,y) in mathbb{R^2} : g(x) leq q leq y} cap {(x,y) in mathbb{R^2} : y leq q leq h(x)}$



      and since $h$ and $g$ are measurable, this is a countable union of measurable sets? I don't feel so good about this reasoning, as I'm pretty sure that I would have to find a way to rewrite $E$ as cartesian product of 2 Lebesgue measurable sets (ideally, one with respect to x and the other with respect to y).
      So perhaps $E=(g^{-1}[-infty,y]cap h^{-1}[y,infty]) times {y}$ works instead?



      As for the second part, I'm not particularly sure how to use Fubini's in this scenario. I'd be very gracious for any help and guidance.






      real-analysis measure-theory lebesgue-integral lebesgue-measure measurable-functions







      share|cite|improve this question















      I have an idea of how to proceed, but I'm suspicious that my efforts were of no use. Let $Ainmathcal{M}$ be Lebesgue measurable, and let $g,h:A rightarrow bar{mathbb{R}}$ be Lebesgue measurable functions with $g(x)leq h(x) forall x in A$. With this, we define $E={(x,y) in mathbb{R^2} : g(x) leq y leq h(x)}$. Prove that $Ein mathcal{M^2}=mathcal{M} times mathcal{M}$.



      And following this, show that if $f:E rightarrow [0,infty]$ is $mathcal{M^2}$ measurable, then
      $displaystyleint_Ef dm^2=int_A(int_{g(x)}^{h(x)}f(x,y) dy) dx$, where $m^2=m times m$.



      For the first part, may we write $E={(x,y) in mathbb{R^2} : g(x) leq y} cap {(x,y) in mathbb{R^2} : y leq h(x)}$
      $Rightarrow E=bigcuplimits_{q in mathbb{Q}} {(x,y) in mathbb{R^2} : g(x) leq q leq y} cap {(x,y) in mathbb{R^2} : y leq q leq h(x)}$



      and since $h$ and $g$ are measurable, this is a countable union of measurable sets? I don't feel so good about this reasoning, as I'm pretty sure that I would have to find a way to rewrite $E$ as cartesian product of 2 Lebesgue measurable sets (ideally, one with respect to x and the other with respect to y).
      So perhaps $E=(g^{-1}[-infty,y]cap h^{-1}[y,infty]) times {y}$ works instead?



      As for the second part, I'm not particularly sure how to use Fubini's in this scenario. I'd be very gracious for any help and guidance.






      real-analysis measure-theory lebesgue-integral lebesgue-measure measurable-functions




      real-analysis measure-theory lebesgue-integral lebesgue-measure measurable-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 22:03

























      asked Dec 1 '18 at 21:14









      DevilofHell'sKitchen

      185




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          It is easier to use that $H colon mathbb{R}^2 rightarrow mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y quad text{and} quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A times mathbb{R}$ to $mathbb{R}$.



          Note that we have used that the composition of measurable functions are measurable and that $$(x,y) rightarrow (g(x),y), quad text{resp.} quad (x,y) rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generater, this already proves the measurability according to the product-$sigma$-algebra.



          Coming back to the first step: We see that the set $$h^{-1}_1([0,infty) cap h_2^{-1}((-infty,0]) = { (x,y) in A times mathbb{R} : g(x) le y le h(x)} $$
          is measurable.



          The second part can be shown by Fubini's theorem as follows: We have
          begin{align}
          int_{mathbb{R}^2} f(x,y) 1_{E}(x,y) , lambda^2(x,y) &= int_{mathbb{R}} int_{mathbb{R}} f(x,y) 1_{E}(x,y) ,dy , dx \
          &=int_{A} int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) ,dy , dx.
          end{align}
          In the last line we have used that $g(x) le h(x)$ for all $x in A$.






          share|cite|improve this answer























          • Very clever use of $h(x,y)$ for the first part! Thank you very much.
            – DevilofHell'sKitchen
            Dec 1 '18 at 22:23








          • 1




            Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
            – Jane Doe
            Dec 2 '18 at 4:05






          • 1




            Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
            – p4sch
            Dec 2 '18 at 10:58










          • @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
            – Jane Doe
            Dec 3 '18 at 20:28










          • We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
            – p4sch
            Dec 4 '18 at 7:24











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          It is easier to use that $H colon mathbb{R}^2 rightarrow mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y quad text{and} quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A times mathbb{R}$ to $mathbb{R}$.



          Note that we have used that the composition of measurable functions are measurable and that $$(x,y) rightarrow (g(x),y), quad text{resp.} quad (x,y) rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generater, this already proves the measurability according to the product-$sigma$-algebra.



          Coming back to the first step: We see that the set $$h^{-1}_1([0,infty) cap h_2^{-1}((-infty,0]) = { (x,y) in A times mathbb{R} : g(x) le y le h(x)} $$
          is measurable.



          The second part can be shown by Fubini's theorem as follows: We have
          begin{align}
          int_{mathbb{R}^2} f(x,y) 1_{E}(x,y) , lambda^2(x,y) &= int_{mathbb{R}} int_{mathbb{R}} f(x,y) 1_{E}(x,y) ,dy , dx \
          &=int_{A} int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) ,dy , dx.
          end{align}
          In the last line we have used that $g(x) le h(x)$ for all $x in A$.






          share|cite|improve this answer























          • Very clever use of $h(x,y)$ for the first part! Thank you very much.
            – DevilofHell'sKitchen
            Dec 1 '18 at 22:23








          • 1




            Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
            – Jane Doe
            Dec 2 '18 at 4:05






          • 1




            Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
            – p4sch
            Dec 2 '18 at 10:58










          • @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
            – Jane Doe
            Dec 3 '18 at 20:28










          • We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
            – p4sch
            Dec 4 '18 at 7:24
















          1














          It is easier to use that $H colon mathbb{R}^2 rightarrow mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y quad text{and} quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A times mathbb{R}$ to $mathbb{R}$.



          Note that we have used that the composition of measurable functions are measurable and that $$(x,y) rightarrow (g(x),y), quad text{resp.} quad (x,y) rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generater, this already proves the measurability according to the product-$sigma$-algebra.



          Coming back to the first step: We see that the set $$h^{-1}_1([0,infty) cap h_2^{-1}((-infty,0]) = { (x,y) in A times mathbb{R} : g(x) le y le h(x)} $$
          is measurable.



          The second part can be shown by Fubini's theorem as follows: We have
          begin{align}
          int_{mathbb{R}^2} f(x,y) 1_{E}(x,y) , lambda^2(x,y) &= int_{mathbb{R}} int_{mathbb{R}} f(x,y) 1_{E}(x,y) ,dy , dx \
          &=int_{A} int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) ,dy , dx.
          end{align}
          In the last line we have used that $g(x) le h(x)$ for all $x in A$.






          share|cite|improve this answer























          • Very clever use of $h(x,y)$ for the first part! Thank you very much.
            – DevilofHell'sKitchen
            Dec 1 '18 at 22:23








          • 1




            Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
            – Jane Doe
            Dec 2 '18 at 4:05






          • 1




            Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
            – p4sch
            Dec 2 '18 at 10:58










          • @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
            – Jane Doe
            Dec 3 '18 at 20:28










          • We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
            – p4sch
            Dec 4 '18 at 7:24














          1












          1








          1






          It is easier to use that $H colon mathbb{R}^2 rightarrow mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y quad text{and} quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A times mathbb{R}$ to $mathbb{R}$.



          Note that we have used that the composition of measurable functions are measurable and that $$(x,y) rightarrow (g(x),y), quad text{resp.} quad (x,y) rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generater, this already proves the measurability according to the product-$sigma$-algebra.



          Coming back to the first step: We see that the set $$h^{-1}_1([0,infty) cap h_2^{-1}((-infty,0]) = { (x,y) in A times mathbb{R} : g(x) le y le h(x)} $$
          is measurable.



          The second part can be shown by Fubini's theorem as follows: We have
          begin{align}
          int_{mathbb{R}^2} f(x,y) 1_{E}(x,y) , lambda^2(x,y) &= int_{mathbb{R}} int_{mathbb{R}} f(x,y) 1_{E}(x,y) ,dy , dx \
          &=int_{A} int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) ,dy , dx.
          end{align}
          In the last line we have used that $g(x) le h(x)$ for all $x in A$.






          share|cite|improve this answer














          It is easier to use that $H colon mathbb{R}^2 rightarrow mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y quad text{and} quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A times mathbb{R}$ to $mathbb{R}$.



          Note that we have used that the composition of measurable functions are measurable and that $$(x,y) rightarrow (g(x),y), quad text{resp.} quad (x,y) rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generater, this already proves the measurability according to the product-$sigma$-algebra.



          Coming back to the first step: We see that the set $$h^{-1}_1([0,infty) cap h_2^{-1}((-infty,0]) = { (x,y) in A times mathbb{R} : g(x) le y le h(x)} $$
          is measurable.



          The second part can be shown by Fubini's theorem as follows: We have
          begin{align}
          int_{mathbb{R}^2} f(x,y) 1_{E}(x,y) , lambda^2(x,y) &= int_{mathbb{R}} int_{mathbb{R}} f(x,y) 1_{E}(x,y) ,dy , dx \
          &=int_{A} int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) ,dy , dx.
          end{align}
          In the last line we have used that $g(x) le h(x)$ for all $x in A$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 10:56

























          answered Dec 1 '18 at 22:10









          p4sch

          4,760217




          4,760217












          • Very clever use of $h(x,y)$ for the first part! Thank you very much.
            – DevilofHell'sKitchen
            Dec 1 '18 at 22:23








          • 1




            Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
            – Jane Doe
            Dec 2 '18 at 4:05






          • 1




            Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
            – p4sch
            Dec 2 '18 at 10:58










          • @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
            – Jane Doe
            Dec 3 '18 at 20:28










          • We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
            – p4sch
            Dec 4 '18 at 7:24


















          • Very clever use of $h(x,y)$ for the first part! Thank you very much.
            – DevilofHell'sKitchen
            Dec 1 '18 at 22:23








          • 1




            Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
            – Jane Doe
            Dec 2 '18 at 4:05






          • 1




            Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
            – p4sch
            Dec 2 '18 at 10:58










          • @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
            – Jane Doe
            Dec 3 '18 at 20:28










          • We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
            – p4sch
            Dec 4 '18 at 7:24
















          Very clever use of $h(x,y)$ for the first part! Thank you very much.
          – DevilofHell'sKitchen
          Dec 1 '18 at 22:23






          Very clever use of $h(x,y)$ for the first part! Thank you very much.
          – DevilofHell'sKitchen
          Dec 1 '18 at 22:23






          1




          1




          Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
          – Jane Doe
          Dec 2 '18 at 4:05




          Just a question: how is it that we are allowed to defined $h$ as you did? It seems to me that you have only showed that the statement holds for your particular choice of $h$. What am I missing?
          – Jane Doe
          Dec 2 '18 at 4:05




          1




          1




          Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
          – p4sch
          Dec 2 '18 at 10:58




          Sorry! I mixed up the notation. $H$ is a auxiliary function - I have changed the notation and call it now with a capital $h$. I have added some more comments in my answer. In particular, why $h_1$ and $h_2$ are measurable.
          – p4sch
          Dec 2 '18 at 10:58












          @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
          – Jane Doe
          Dec 3 '18 at 20:28




          @p4sch thanks for the clarification. The only thing I’m still confused about is, how do we rid ourselves of the characteristic function in the integrand (on the last line)? It’s clear to me why the first one is the integral over $E$. Does this just follow from the fact that the integral evaluates to $0$ off $E$?
          – Jane Doe
          Dec 3 '18 at 20:28












          We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
          – p4sch
          Dec 4 '18 at 7:24




          We have by definition $(x,y) in E$ if and only if $x in A$ and $y in [g(x),h(x)]$. Thus, I have only rewritten the last integral.
          – p4sch
          Dec 4 '18 at 7:24


















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