Tightness of a finite measure on a complete and separable metric space












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Let $(X, rho)$ be a metric spac and let $mu$ be a finite Borel measure on it. Assume thay $(X, rho)$ is complete and separable. Prove that for every $epsilon > 0$ there is a complact subset $K subset X$ such that $mu(K^c) < epsilon$.




I am not sure what to do here. We are to use that a subset of a complete metric space is compact iff it is closed and totally bounded. I managed to prove that for any $delta > 0$ there is a finite subset ${x_1,...,x_n}$ such that $mu((B(x_1,delta) cup... cup B(x_n,delta))^c) < epsilon$. It seems like I have to take something like closure, but I am not sure how.






real-analysis measure-theory compactness complete-spaces







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    Let $(X, rho)$ be a metric spac and let $mu$ be a finite Borel measure on it. Assume thay $(X, rho)$ is complete and separable. Prove that for every $epsilon > 0$ there is a complact subset $K subset X$ such that $mu(K^c) < epsilon$.




    I am not sure what to do here. We are to use that a subset of a complete metric space is compact iff it is closed and totally bounded. I managed to prove that for any $delta > 0$ there is a finite subset ${x_1,...,x_n}$ such that $mu((B(x_1,delta) cup... cup B(x_n,delta))^c) < epsilon$. It seems like I have to take something like closure, but I am not sure how.






    real-analysis measure-theory compactness complete-spaces







    share|cite|improve this question

























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      Let $(X, rho)$ be a metric spac and let $mu$ be a finite Borel measure on it. Assume thay $(X, rho)$ is complete and separable. Prove that for every $epsilon > 0$ there is a complact subset $K subset X$ such that $mu(K^c) < epsilon$.




      I am not sure what to do here. We are to use that a subset of a complete metric space is compact iff it is closed and totally bounded. I managed to prove that for any $delta > 0$ there is a finite subset ${x_1,...,x_n}$ such that $mu((B(x_1,delta) cup... cup B(x_n,delta))^c) < epsilon$. It seems like I have to take something like closure, but I am not sure how.






      real-analysis measure-theory compactness complete-spaces







      share|cite|improve this question














      Let $(X, rho)$ be a metric spac and let $mu$ be a finite Borel measure on it. Assume thay $(X, rho)$ is complete and separable. Prove that for every $epsilon > 0$ there is a complact subset $K subset X$ such that $mu(K^c) < epsilon$.




      I am not sure what to do here. We are to use that a subset of a complete metric space is compact iff it is closed and totally bounded. I managed to prove that for any $delta > 0$ there is a finite subset ${x_1,...,x_n}$ such that $mu((B(x_1,delta) cup... cup B(x_n,delta))^c) < epsilon$. It seems like I have to take something like closure, but I am not sure how.






      real-analysis measure-theory compactness complete-spaces




      real-analysis measure-theory compactness complete-spaces






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      asked Nov 25 at 4:46









      Cute Brownie

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          For each $n$ take $delta =frac 1 n$ and replace $epsilon$ by $epsilon /2^{n}$ in your argument to get a finite set ${x_{n1},x_{n2},...,x_{nk_n}}$. Consider $H=cap_n [B(x_{n1},frac 1 n)cup B(x_{n2},frac 1 n),...,cup B(x_{nk_n},frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $epsilon$. Hence $mu(K^{c}) <epsilon$.






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            For each $n$ take $delta =frac 1 n$ and replace $epsilon$ by $epsilon /2^{n}$ in your argument to get a finite set ${x_{n1},x_{n2},...,x_{nk_n}}$. Consider $H=cap_n [B(x_{n1},frac 1 n)cup B(x_{n2},frac 1 n),...,cup B(x_{nk_n},frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $epsilon$. Hence $mu(K^{c}) <epsilon$.






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              For each $n$ take $delta =frac 1 n$ and replace $epsilon$ by $epsilon /2^{n}$ in your argument to get a finite set ${x_{n1},x_{n2},...,x_{nk_n}}$. Consider $H=cap_n [B(x_{n1},frac 1 n)cup B(x_{n2},frac 1 n),...,cup B(x_{nk_n},frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $epsilon$. Hence $mu(K^{c}) <epsilon$.






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                For each $n$ take $delta =frac 1 n$ and replace $epsilon$ by $epsilon /2^{n}$ in your argument to get a finite set ${x_{n1},x_{n2},...,x_{nk_n}}$. Consider $H=cap_n [B(x_{n1},frac 1 n)cup B(x_{n2},frac 1 n),...,cup B(x_{nk_n},frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $epsilon$. Hence $mu(K^{c}) <epsilon$.






                share|cite|improve this answer












                For each $n$ take $delta =frac 1 n$ and replace $epsilon$ by $epsilon /2^{n}$ in your argument to get a finite set ${x_{n1},x_{n2},...,x_{nk_n}}$. Consider $H=cap_n [B(x_{n1},frac 1 n)cup B(x_{n2},frac 1 n),...,cup B(x_{nk_n},frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $epsilon$. Hence $mu(K^{c}) <epsilon$.







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                answered Nov 25 at 4:56









                Kavi Rama Murthy

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