Weak convergence in the Skorohod space $D([0, T])$ $forall T$ implies Weak convergence in $D([0, infty))$?












3














Assume that $Z_n$ are random variables taking value in the Skorohod space $D([0, infty),Y)$ (endowed with its usual Skorohod topology) of right-continuous functions $[0, infty) to Y$, where $Y$ is a real separable Banach space. Assume that $forall T>0$, $Z_n$ converge weakly (in distribution) to $Z$ in the space $D([0, T],Y)$, where $Z$ is a random variable defined on $D([0, infty),Y)$. Does it hold that $Z_n$ converge weakly to $Z$ in the space $D([0, infty),Y)$? What about if an assumption is made on $Z$, for example $Z$ is a constant random variable or $Z$ is continuous, i.e. it belongs to $C([0, infty),Y)$.






probability-theory convergence random-variables weak-convergence







share|cite|improve this question
























  • the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
    – user44670
    Aug 23 '13 at 17:21












  • How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
    – Davide Giraudo
    Aug 23 '13 at 17:25










  • Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
    – user44670
    Aug 23 '13 at 17:31












  • so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
    – user44670
    Aug 23 '13 at 17:34










  • I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
    – Davide Giraudo
    Aug 23 '13 at 19:49


















3














Assume that $Z_n$ are random variables taking value in the Skorohod space $D([0, infty),Y)$ (endowed with its usual Skorohod topology) of right-continuous functions $[0, infty) to Y$, where $Y$ is a real separable Banach space. Assume that $forall T>0$, $Z_n$ converge weakly (in distribution) to $Z$ in the space $D([0, T],Y)$, where $Z$ is a random variable defined on $D([0, infty),Y)$. Does it hold that $Z_n$ converge weakly to $Z$ in the space $D([0, infty),Y)$? What about if an assumption is made on $Z$, for example $Z$ is a constant random variable or $Z$ is continuous, i.e. it belongs to $C([0, infty),Y)$.






probability-theory convergence random-variables weak-convergence







share|cite|improve this question
























  • the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
    – user44670
    Aug 23 '13 at 17:21












  • How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
    – Davide Giraudo
    Aug 23 '13 at 17:25










  • Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
    – user44670
    Aug 23 '13 at 17:31












  • so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
    – user44670
    Aug 23 '13 at 17:34










  • I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
    – Davide Giraudo
    Aug 23 '13 at 19:49
















3












3








3


1





Assume that $Z_n$ are random variables taking value in the Skorohod space $D([0, infty),Y)$ (endowed with its usual Skorohod topology) of right-continuous functions $[0, infty) to Y$, where $Y$ is a real separable Banach space. Assume that $forall T>0$, $Z_n$ converge weakly (in distribution) to $Z$ in the space $D([0, T],Y)$, where $Z$ is a random variable defined on $D([0, infty),Y)$. Does it hold that $Z_n$ converge weakly to $Z$ in the space $D([0, infty),Y)$? What about if an assumption is made on $Z$, for example $Z$ is a constant random variable or $Z$ is continuous, i.e. it belongs to $C([0, infty),Y)$.






probability-theory convergence random-variables weak-convergence







share|cite|improve this question















Assume that $Z_n$ are random variables taking value in the Skorohod space $D([0, infty),Y)$ (endowed with its usual Skorohod topology) of right-continuous functions $[0, infty) to Y$, where $Y$ is a real separable Banach space. Assume that $forall T>0$, $Z_n$ converge weakly (in distribution) to $Z$ in the space $D([0, T],Y)$, where $Z$ is a random variable defined on $D([0, infty),Y)$. Does it hold that $Z_n$ converge weakly to $Z$ in the space $D([0, infty),Y)$? What about if an assumption is made on $Z$, for example $Z$ is a constant random variable or $Z$ is continuous, i.e. it belongs to $C([0, infty),Y)$.






probability-theory convergence random-variables weak-convergence




probability-theory convergence random-variables weak-convergence






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share|cite|improve this question













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edited Aug 22 '13 at 17:36

























asked Aug 19 '13 at 0:18









user44670

13011




13011












  • the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
    – user44670
    Aug 23 '13 at 17:21












  • How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
    – Davide Giraudo
    Aug 23 '13 at 17:25










  • Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
    – user44670
    Aug 23 '13 at 17:31












  • so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
    – user44670
    Aug 23 '13 at 17:34










  • I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
    – Davide Giraudo
    Aug 23 '13 at 19:49




















  • the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
    – user44670
    Aug 23 '13 at 17:21












  • How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
    – Davide Giraudo
    Aug 23 '13 at 17:25










  • Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
    – user44670
    Aug 23 '13 at 17:31












  • so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
    – user44670
    Aug 23 '13 at 17:34










  • I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
    – Davide Giraudo
    Aug 23 '13 at 19:49


















the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
– user44670
Aug 23 '13 at 17:21






the metric on $D[0,infty)$ is something like $int_0^infty e^{-t} sup_{u leq t} ||z_n(u)-z(u)||dt$. Therefore if you take your example of a continuous function equals to 1 on $[0,n]$ and 0 after $n+1$, then we have convergence to $z=1$ on $D[0,T]$ and also on $D[0,infty)$ because $d(z_n,z)leq e^{-n} to 0$.
– user44670
Aug 23 '13 at 17:21














How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
– Davide Giraudo
Aug 23 '13 at 17:25




How can you be sure that this metric is well-defined (for example with $Y=mathbb R$ and $z(u):=e^{u^2}$)? Do you have to assume boundedness?
– Davide Giraudo
Aug 23 '13 at 17:25












Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
– user44670
Aug 23 '13 at 17:31






Good remark. The answer is because I did not give you the exact metric, I said "something like", to have the spirit. The real metric is: $d(z_n,z)= inf_{lambda in Lambda} [gamma(lambda) vee int_0^infty e^{-t} [1 wedge (sup_{u leq t} ||z_n(t)-z(lambda(t))||)]dt ]$, where $Lambda$ is the set of strictly increasing lipschitz continuous functions, with $lambda(0)=0$.
– user44670
Aug 23 '13 at 17:31














so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
– user44670
Aug 23 '13 at 17:34




so the answer to your question is that the $sup$ is bounded by 1 inside the integral.
– user44670
Aug 23 '13 at 17:34












I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
– Davide Giraudo
Aug 23 '13 at 19:49






I see. So in this context, convergence of finite-dimensional distributions is not a problem. Have you checked tightness criterion?
– Davide Giraudo
Aug 23 '13 at 19:49












1 Answer
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oldest

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We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):




Theorem. Let $(S,d)$ be a separable metric space and for $u,ninmathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $left(Omega,mathcal F,muright)$. We assume that




  1. for all $uinmathbb N$, $X_{un}to X_u$ in distribution as $nto infty$;


  2. $X_uto X$ in distribution as $uto infty$, and

  3. for each $varepsilongt 0$, $lim_{uto infty}limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}=0$.


Then $Y_nto X$ in distribution as $nto infty$.




The proof uses portmanteau theorem: we take $Fsubset S$ a closed set, and we define $F_varepsilon:={x,d(x,F)leqslantvarepsilon}$, where $d(x,F):=inf{d(x,y),yin F}$. Then
$$mu{Y_nin F}leqslant mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon},$$
hence taking the $limsup_{nto infty}$,
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{X_uin F_varepsilon}+limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}.$$
Taking now $lim_{uto infty}$, we get
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{Xin F_varepsilon},$$
and we conclude, using the fact that $bigcap_{varepsilon >0}F_varepsilon=F$, $F$ begin closed.



Now, we shall use this result to prove that if $Z_nto Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,infty)$. We define
$$X_{u,n}(t,omega):=Z_n(t,omega)chi_{[0,u)}(t)$$
$$Y_n:=Z_n; quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,infty)$, we have that
$$d(X_{u,n},Y_n)leqslant int_u^{infty}e^{-t}mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $gamma$ for which $gamma(s)=s$ for $0leqslant sleqslant u$.






share|cite|improve this answer























  • Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
    – 0xbadf00d
    Dec 1 at 21:45










  • @0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
    – Davide Giraudo
    Dec 1 at 22:01










  • Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
    – 0xbadf00d
    Dec 1 at 23:22











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):




Theorem. Let $(S,d)$ be a separable metric space and for $u,ninmathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $left(Omega,mathcal F,muright)$. We assume that




  1. for all $uinmathbb N$, $X_{un}to X_u$ in distribution as $nto infty$;


  2. $X_uto X$ in distribution as $uto infty$, and

  3. for each $varepsilongt 0$, $lim_{uto infty}limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}=0$.


Then $Y_nto X$ in distribution as $nto infty$.




The proof uses portmanteau theorem: we take $Fsubset S$ a closed set, and we define $F_varepsilon:={x,d(x,F)leqslantvarepsilon}$, where $d(x,F):=inf{d(x,y),yin F}$. Then
$$mu{Y_nin F}leqslant mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon},$$
hence taking the $limsup_{nto infty}$,
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{X_uin F_varepsilon}+limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}.$$
Taking now $lim_{uto infty}$, we get
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{Xin F_varepsilon},$$
and we conclude, using the fact that $bigcap_{varepsilon >0}F_varepsilon=F$, $F$ begin closed.



Now, we shall use this result to prove that if $Z_nto Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,infty)$. We define
$$X_{u,n}(t,omega):=Z_n(t,omega)chi_{[0,u)}(t)$$
$$Y_n:=Z_n; quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,infty)$, we have that
$$d(X_{u,n},Y_n)leqslant int_u^{infty}e^{-t}mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $gamma$ for which $gamma(s)=s$ for $0leqslant sleqslant u$.






share|cite|improve this answer























  • Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
    – 0xbadf00d
    Dec 1 at 21:45










  • @0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
    – Davide Giraudo
    Dec 1 at 22:01










  • Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
    – 0xbadf00d
    Dec 1 at 23:22
















2














We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):




Theorem. Let $(S,d)$ be a separable metric space and for $u,ninmathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $left(Omega,mathcal F,muright)$. We assume that




  1. for all $uinmathbb N$, $X_{un}to X_u$ in distribution as $nto infty$;


  2. $X_uto X$ in distribution as $uto infty$, and

  3. for each $varepsilongt 0$, $lim_{uto infty}limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}=0$.


Then $Y_nto X$ in distribution as $nto infty$.




The proof uses portmanteau theorem: we take $Fsubset S$ a closed set, and we define $F_varepsilon:={x,d(x,F)leqslantvarepsilon}$, where $d(x,F):=inf{d(x,y),yin F}$. Then
$$mu{Y_nin F}leqslant mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon},$$
hence taking the $limsup_{nto infty}$,
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{X_uin F_varepsilon}+limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}.$$
Taking now $lim_{uto infty}$, we get
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{Xin F_varepsilon},$$
and we conclude, using the fact that $bigcap_{varepsilon >0}F_varepsilon=F$, $F$ begin closed.



Now, we shall use this result to prove that if $Z_nto Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,infty)$. We define
$$X_{u,n}(t,omega):=Z_n(t,omega)chi_{[0,u)}(t)$$
$$Y_n:=Z_n; quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,infty)$, we have that
$$d(X_{u,n},Y_n)leqslant int_u^{infty}e^{-t}mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $gamma$ for which $gamma(s)=s$ for $0leqslant sleqslant u$.






share|cite|improve this answer























  • Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
    – 0xbadf00d
    Dec 1 at 21:45










  • @0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
    – Davide Giraudo
    Dec 1 at 22:01










  • Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
    – 0xbadf00d
    Dec 1 at 23:22














2












2








2






We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):




Theorem. Let $(S,d)$ be a separable metric space and for $u,ninmathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $left(Omega,mathcal F,muright)$. We assume that




  1. for all $uinmathbb N$, $X_{un}to X_u$ in distribution as $nto infty$;


  2. $X_uto X$ in distribution as $uto infty$, and

  3. for each $varepsilongt 0$, $lim_{uto infty}limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}=0$.


Then $Y_nto X$ in distribution as $nto infty$.




The proof uses portmanteau theorem: we take $Fsubset S$ a closed set, and we define $F_varepsilon:={x,d(x,F)leqslantvarepsilon}$, where $d(x,F):=inf{d(x,y),yin F}$. Then
$$mu{Y_nin F}leqslant mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon},$$
hence taking the $limsup_{nto infty}$,
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{X_uin F_varepsilon}+limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}.$$
Taking now $lim_{uto infty}$, we get
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{Xin F_varepsilon},$$
and we conclude, using the fact that $bigcap_{varepsilon >0}F_varepsilon=F$, $F$ begin closed.



Now, we shall use this result to prove that if $Z_nto Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,infty)$. We define
$$X_{u,n}(t,omega):=Z_n(t,omega)chi_{[0,u)}(t)$$
$$Y_n:=Z_n; quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,infty)$, we have that
$$d(X_{u,n},Y_n)leqslant int_u^{infty}e^{-t}mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $gamma$ for which $gamma(s)=s$ for $0leqslant sleqslant u$.






share|cite|improve this answer














We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):




Theorem. Let $(S,d)$ be a separable metric space and for $u,ninmathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $left(Omega,mathcal F,muright)$. We assume that




  1. for all $uinmathbb N$, $X_{un}to X_u$ in distribution as $nto infty$;


  2. $X_uto X$ in distribution as $uto infty$, and

  3. for each $varepsilongt 0$, $lim_{uto infty}limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}=0$.


Then $Y_nto X$ in distribution as $nto infty$.




The proof uses portmanteau theorem: we take $Fsubset S$ a closed set, and we define $F_varepsilon:={x,d(x,F)leqslantvarepsilon}$, where $d(x,F):=inf{d(x,y),yin F}$. Then
$$mu{Y_nin F}leqslant mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon},$$
hence taking the $limsup_{nto infty}$,
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{X_uin F_varepsilon}+limsup_{nto infty}mu{d(X_{u,n},Y_n)gtvarepsilon}.$$
Taking now $lim_{uto infty}$, we get
$$limsup_{nto infty}mu{Y_nin F}leqslant mu{Xin F_varepsilon},$$
and we conclude, using the fact that $bigcap_{varepsilon >0}F_varepsilon=F$, $F$ begin closed.



Now, we shall use this result to prove that if $Z_nto Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,infty)$. We define
$$X_{u,n}(t,omega):=Z_n(t,omega)chi_{[0,u)}(t)$$
$$Y_n:=Z_n; quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,infty)$, we have that
$$d(X_{u,n},Y_n)leqslant int_u^{infty}e^{-t}mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $gamma$ for which $gamma(s)=s$ for $0leqslant sleqslant u$.







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edited Dec 1 at 22:00

























answered Aug 24 '13 at 12:14









Davide Giraudo

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  • Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
    – 0xbadf00d
    Dec 1 at 21:45










  • @0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
    – Davide Giraudo
    Dec 1 at 22:01










  • Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
    – 0xbadf00d
    Dec 1 at 23:22


















  • Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
    – 0xbadf00d
    Dec 1 at 21:45










  • @0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
    – Davide Giraudo
    Dec 1 at 22:01










  • Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
    – 0xbadf00d
    Dec 1 at 23:22
















Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
– 0xbadf00d
Dec 1 at 21:45




Sorry for necroing this old answer, but how did you obtain $mu{Y_nin F}=mu{X_{u,n}in F_varepsilon}+mu{d(X_{u,n},Y_n)gtvarepsilon}$?
– 0xbadf00d
Dec 1 at 21:45












@0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
– Davide Giraudo
Dec 1 at 22:01




@0xbadf00d It was supposed to be an inequality. Thanks for pointing this out.
– Davide Giraudo
Dec 1 at 22:01












Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
– 0xbadf00d
Dec 1 at 23:22




Thought I would miss something. Thanks for the quick response. I've got a question where a theorem similar the one you've proven here is needed. Maybe you can help there too.
– 0xbadf00d
Dec 1 at 23:22


















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