Egorov's theorem on intervals











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Frm Folland's Real Analysis exercise 2.43 part (b):




Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.




Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.



I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).






real-analysis measure-theory uniform-convergence







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    Frm Folland's Real Analysis exercise 2.43 part (b):




    Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.




    Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.



    I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).






    real-analysis measure-theory uniform-convergence







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Frm Folland's Real Analysis exercise 2.43 part (b):




      Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.




      Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.



      I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).






      real-analysis measure-theory uniform-convergence







      share|cite|improve this question















      Frm Folland's Real Analysis exercise 2.43 part (b):




      Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.




      Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.



      I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).






      real-analysis measure-theory uniform-convergence




      real-analysis measure-theory uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 24 at 6:32

























      asked Nov 24 at 6:06









      Cute Brownie

      990416




      990416






















          1 Answer
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          up vote
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          accepted










          Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
          begin{align*}
          bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
          end{align*}
          Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
          begin{align*}
          mu(A_{m,k})<frac{epsilon}{2^m}.
          end{align*}
          Thus,
          begin{align}
          E:=bigcup_{min mathbb{N}} A_{m,k}
          end{align}
          has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.






          share|cite|improve this answer























          • Yea, this solution is correct. +1
            – mathworker21
            Nov 24 at 6:50










          • Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
            – Cute Brownie
            Nov 24 at 7:03










          • Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
            – confused_wallet
            Nov 24 at 7:05











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          up vote
          2
          down vote



          accepted










          Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
          begin{align*}
          bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
          end{align*}
          Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
          begin{align*}
          mu(A_{m,k})<frac{epsilon}{2^m}.
          end{align*}
          Thus,
          begin{align}
          E:=bigcup_{min mathbb{N}} A_{m,k}
          end{align}
          has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.






          share|cite|improve this answer























          • Yea, this solution is correct. +1
            – mathworker21
            Nov 24 at 6:50










          • Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
            – Cute Brownie
            Nov 24 at 7:03










          • Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
            – confused_wallet
            Nov 24 at 7:05















          up vote
          2
          down vote



          accepted










          Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
          begin{align*}
          bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
          end{align*}
          Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
          begin{align*}
          mu(A_{m,k})<frac{epsilon}{2^m}.
          end{align*}
          Thus,
          begin{align}
          E:=bigcup_{min mathbb{N}} A_{m,k}
          end{align}
          has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.






          share|cite|improve this answer























          • Yea, this solution is correct. +1
            – mathworker21
            Nov 24 at 6:50










          • Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
            – Cute Brownie
            Nov 24 at 7:03










          • Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
            – confused_wallet
            Nov 24 at 7:05













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
          begin{align*}
          bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
          end{align*}
          Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
          begin{align*}
          mu(A_{m,k})<frac{epsilon}{2^m}.
          end{align*}
          Thus,
          begin{align}
          E:=bigcup_{min mathbb{N}} A_{m,k}
          end{align}
          has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.






          share|cite|improve this answer














          Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
          begin{align*}
          bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
          end{align*}
          Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
          begin{align*}
          mu(A_{m,k})<frac{epsilon}{2^m}.
          end{align*}
          Thus,
          begin{align}
          E:=bigcup_{min mathbb{N}} A_{m,k}
          end{align}
          has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 6:57

























          answered Nov 24 at 6:28









          confused_wallet

          473311




          473311












          • Yea, this solution is correct. +1
            – mathworker21
            Nov 24 at 6:50










          • Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
            – Cute Brownie
            Nov 24 at 7:03










          • Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
            – confused_wallet
            Nov 24 at 7:05


















          • Yea, this solution is correct. +1
            – mathworker21
            Nov 24 at 6:50










          • Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
            – Cute Brownie
            Nov 24 at 7:03










          • Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
            – confused_wallet
            Nov 24 at 7:05
















          Yea, this solution is correct. +1
          – mathworker21
          Nov 24 at 6:50




          Yea, this solution is correct. +1
          – mathworker21
          Nov 24 at 6:50












          Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
          – Cute Brownie
          Nov 24 at 7:03




          Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
          – Cute Brownie
          Nov 24 at 7:03












          Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
          – confused_wallet
          Nov 24 at 7:05




          Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
          – confused_wallet
          Nov 24 at 7:05


















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