Is the domain of convergence of a sequence of measurable functions measurable? (general targets)











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Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
$$ (X,Sigma) to (Y,B(Y)).$$




Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?




This is well-known (and easy) in the case of real-valued functions. Indeed,



$$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$



I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
$$ dleft(f_{r}(x),f_{s}(x)right) .$$



Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?










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    up vote
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    down vote

    favorite
    4












    Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
    Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
    $$ (X,Sigma) to (Y,B(Y)).$$




    Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?




    This is well-known (and easy) in the case of real-valued functions. Indeed,



    $$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$



    I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
    $$ dleft(f_{r}(x),f_{s}(x)right) .$$



    Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?










    share|cite|improve this question


























      up vote
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      down vote

      favorite
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      up vote
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      4





      Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
      Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
      $$ (X,Sigma) to (Y,B(Y)).$$




      Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?




      This is well-known (and easy) in the case of real-valued functions. Indeed,



      $$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$



      I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
      $$ dleft(f_{r}(x),f_{s}(x)right) .$$



      Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?










      share|cite|improve this question















      Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
      Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
      $$ (X,Sigma) to (Y,B(Y)).$$




      Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?




      This is well-known (and easy) in the case of real-valued functions. Indeed,



      $$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$



      I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
      $$ dleft(f_{r}(x),f_{s}(x)right) .$$



      Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?







      measure-theory measurable-functions






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      edited Nov 21 at 14:58









      A. Pongrácz

      5,168725




      5,168725










      asked Nov 21 at 14:48









      Asaf Shachar

      4,9303939




      4,9303939






















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          This is a really instructive question: apparently, completeness is the key!



          First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.



          Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
          Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
          In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
          It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.



          We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
          Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!



          Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
          However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
          As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).






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            up vote
            5
            down vote



            accepted










            This is a really instructive question: apparently, completeness is the key!



            First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.



            Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
            Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
            In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
            It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.



            We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
            Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!



            Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
            However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
            As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).






            share|cite|improve this answer



























              up vote
              5
              down vote



              accepted










              This is a really instructive question: apparently, completeness is the key!



              First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.



              Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
              Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
              In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
              It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.



              We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
              Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!



              Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
              However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
              As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).






              share|cite|improve this answer

























                up vote
                5
                down vote



                accepted







                up vote
                5
                down vote



                accepted






                This is a really instructive question: apparently, completeness is the key!



                First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.



                Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
                Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
                In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
                It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.



                We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
                Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!



                Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
                However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
                As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).






                share|cite|improve this answer














                This is a really instructive question: apparently, completeness is the key!



                First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.



                Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
                Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
                In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
                It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.



                We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
                Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!



                Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
                However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
                As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 21 at 21:06

























                answered Nov 21 at 20:51









                A. Pongrácz

                5,168725




                5,168725






























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