Continuity from below of outer measure extending an algebrea











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Consider the following problem:




Let $mathcal{A}$ be an Algebra with an additive function $mu$ and let $mu^*$ be the outer measure corresponding to $mu$. Show that if $A_nuparrow A$ then $mu^*(A_n)rightarrow mu^*(A)$.




After much effort I've crafted this attempt at a solution:




For every $n$ we have a cover by sets from
$mathcal{A}:space{P_{n,i}}^{infty}_{i=1}$ such that
$sum_{i}mu(P_{n,i})<mu^*(A_n)+epsilon$. We then also have a $k_n$
such that:
$$sum_{i=k_n+1}^{infty}mu(P_{n,i})<frac{epsilon}{2^n}$$ We will
also define $$R_n:=bigcup_{i=1}^{k_n} P_{n,i}inmathcal{A}$$ We then
have:
$$Asubseteqbigcup_{n=1}^{infty}(R_nsetminusbigcup_{j=1}^{n-1}R_j)cupbigcup_{n=1}^{infty}bigcup_{i=k_n+1}^{infty}P_{n,i}$$
And so by additivity of $mu$:
$$mu^*(A)leqsum_{n=1}^{infty}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)+sum_{n=1}^{infty}sum_{i=k_n+1}^{infty}mu(P_{n.i})leq
epsilon +
lim_msum_{n=1}^{m}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)=\
epsilon+lim_mmu(bigcup_{n=1}^{m}R_n)$$




This is where I'm stuck. Essentially I want to say that the union above isn't much larger than $R_m$ because $R_m$ covers most of $A_n$ for $nlt m$, and $R_n$ isn't much larger than $A_n$, but I'm not sure this can actually be done.



Is this solution salvageable? Or is there another approach that might work better?










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    Consider the following problem:




    Let $mathcal{A}$ be an Algebra with an additive function $mu$ and let $mu^*$ be the outer measure corresponding to $mu$. Show that if $A_nuparrow A$ then $mu^*(A_n)rightarrow mu^*(A)$.




    After much effort I've crafted this attempt at a solution:




    For every $n$ we have a cover by sets from
    $mathcal{A}:space{P_{n,i}}^{infty}_{i=1}$ such that
    $sum_{i}mu(P_{n,i})<mu^*(A_n)+epsilon$. We then also have a $k_n$
    such that:
    $$sum_{i=k_n+1}^{infty}mu(P_{n,i})<frac{epsilon}{2^n}$$ We will
    also define $$R_n:=bigcup_{i=1}^{k_n} P_{n,i}inmathcal{A}$$ We then
    have:
    $$Asubseteqbigcup_{n=1}^{infty}(R_nsetminusbigcup_{j=1}^{n-1}R_j)cupbigcup_{n=1}^{infty}bigcup_{i=k_n+1}^{infty}P_{n,i}$$
    And so by additivity of $mu$:
    $$mu^*(A)leqsum_{n=1}^{infty}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)+sum_{n=1}^{infty}sum_{i=k_n+1}^{infty}mu(P_{n.i})leq
    epsilon +
    lim_msum_{n=1}^{m}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)=\
    epsilon+lim_mmu(bigcup_{n=1}^{m}R_n)$$




    This is where I'm stuck. Essentially I want to say that the union above isn't much larger than $R_m$ because $R_m$ covers most of $A_n$ for $nlt m$, and $R_n$ isn't much larger than $A_n$, but I'm not sure this can actually be done.



    Is this solution salvageable? Or is there another approach that might work better?










    share|cite|improve this question
























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      favorite
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      up vote
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      Consider the following problem:




      Let $mathcal{A}$ be an Algebra with an additive function $mu$ and let $mu^*$ be the outer measure corresponding to $mu$. Show that if $A_nuparrow A$ then $mu^*(A_n)rightarrow mu^*(A)$.




      After much effort I've crafted this attempt at a solution:




      For every $n$ we have a cover by sets from
      $mathcal{A}:space{P_{n,i}}^{infty}_{i=1}$ such that
      $sum_{i}mu(P_{n,i})<mu^*(A_n)+epsilon$. We then also have a $k_n$
      such that:
      $$sum_{i=k_n+1}^{infty}mu(P_{n,i})<frac{epsilon}{2^n}$$ We will
      also define $$R_n:=bigcup_{i=1}^{k_n} P_{n,i}inmathcal{A}$$ We then
      have:
      $$Asubseteqbigcup_{n=1}^{infty}(R_nsetminusbigcup_{j=1}^{n-1}R_j)cupbigcup_{n=1}^{infty}bigcup_{i=k_n+1}^{infty}P_{n,i}$$
      And so by additivity of $mu$:
      $$mu^*(A)leqsum_{n=1}^{infty}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)+sum_{n=1}^{infty}sum_{i=k_n+1}^{infty}mu(P_{n.i})leq
      epsilon +
      lim_msum_{n=1}^{m}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)=\
      epsilon+lim_mmu(bigcup_{n=1}^{m}R_n)$$




      This is where I'm stuck. Essentially I want to say that the union above isn't much larger than $R_m$ because $R_m$ covers most of $A_n$ for $nlt m$, and $R_n$ isn't much larger than $A_n$, but I'm not sure this can actually be done.



      Is this solution salvageable? Or is there another approach that might work better?










      share|cite|improve this question













      Consider the following problem:




      Let $mathcal{A}$ be an Algebra with an additive function $mu$ and let $mu^*$ be the outer measure corresponding to $mu$. Show that if $A_nuparrow A$ then $mu^*(A_n)rightarrow mu^*(A)$.




      After much effort I've crafted this attempt at a solution:




      For every $n$ we have a cover by sets from
      $mathcal{A}:space{P_{n,i}}^{infty}_{i=1}$ such that
      $sum_{i}mu(P_{n,i})<mu^*(A_n)+epsilon$. We then also have a $k_n$
      such that:
      $$sum_{i=k_n+1}^{infty}mu(P_{n,i})<frac{epsilon}{2^n}$$ We will
      also define $$R_n:=bigcup_{i=1}^{k_n} P_{n,i}inmathcal{A}$$ We then
      have:
      $$Asubseteqbigcup_{n=1}^{infty}(R_nsetminusbigcup_{j=1}^{n-1}R_j)cupbigcup_{n=1}^{infty}bigcup_{i=k_n+1}^{infty}P_{n,i}$$
      And so by additivity of $mu$:
      $$mu^*(A)leqsum_{n=1}^{infty}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)+sum_{n=1}^{infty}sum_{i=k_n+1}^{infty}mu(P_{n.i})leq
      epsilon +
      lim_msum_{n=1}^{m}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)=\
      epsilon+lim_mmu(bigcup_{n=1}^{m}R_n)$$




      This is where I'm stuck. Essentially I want to say that the union above isn't much larger than $R_m$ because $R_m$ covers most of $A_n$ for $nlt m$, and $R_n$ isn't much larger than $A_n$, but I'm not sure this can actually be done.



      Is this solution salvageable? Or is there another approach that might work better?







      real-analysis measure-theory outer-measure






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      asked Nov 14 at 17:26









      Bar Alon

      465114




      465114






















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          This is only a partial answer. I am not sure, if the statement is in general true!



          If the additive measure $mu$ on $mathcal{A}$ has the property that it extends to a measure $widetilde{mu}$ on a $sigma$-algebra $mathcal{F} supset mathcal{A}$ with the property that for any $A subset X$ there exists a $B in mathcal{F}$ with $A subset B$ and
          $$mu^*(A) = widetilde{mu}(B),$$
          then the statement is true. For example, this is the case if $mu$ is the Lebesgue-measure on $mathbb{R}$ and $mathcal{A}$ is the $sigma$-algebra of all Borel-sets.



          Proof: Take for $A_n$ the corresponding $B_n in mathcal{F}$ with $mu^*(A_n) = widetilde{mu}(B_n)$. Then $x in A$ implies that $x in A_m$ for all $m ge n$, starting at some $n in mathbb{N}$. Thus $x in liminf_{n rightarrow infty} B_n$, because $A_n subset B_n$, and
          $$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$






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          • Can you clarify your notation a bit? What are $G_n$? and what is $E$?
            – Bar Alon
            Nov 16 at 8:38










          • Sorry, I have corrected the typos: $G_n$ should be $B_n$.
            – p4sch
            Nov 16 at 8:55











          Your Answer





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          up vote
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          This is only a partial answer. I am not sure, if the statement is in general true!



          If the additive measure $mu$ on $mathcal{A}$ has the property that it extends to a measure $widetilde{mu}$ on a $sigma$-algebra $mathcal{F} supset mathcal{A}$ with the property that for any $A subset X$ there exists a $B in mathcal{F}$ with $A subset B$ and
          $$mu^*(A) = widetilde{mu}(B),$$
          then the statement is true. For example, this is the case if $mu$ is the Lebesgue-measure on $mathbb{R}$ and $mathcal{A}$ is the $sigma$-algebra of all Borel-sets.



          Proof: Take for $A_n$ the corresponding $B_n in mathcal{F}$ with $mu^*(A_n) = widetilde{mu}(B_n)$. Then $x in A$ implies that $x in A_m$ for all $m ge n$, starting at some $n in mathbb{N}$. Thus $x in liminf_{n rightarrow infty} B_n$, because $A_n subset B_n$, and
          $$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$






          share|cite|improve this answer























          • Can you clarify your notation a bit? What are $G_n$? and what is $E$?
            – Bar Alon
            Nov 16 at 8:38










          • Sorry, I have corrected the typos: $G_n$ should be $B_n$.
            – p4sch
            Nov 16 at 8:55















          up vote
          1
          down vote













          This is only a partial answer. I am not sure, if the statement is in general true!



          If the additive measure $mu$ on $mathcal{A}$ has the property that it extends to a measure $widetilde{mu}$ on a $sigma$-algebra $mathcal{F} supset mathcal{A}$ with the property that for any $A subset X$ there exists a $B in mathcal{F}$ with $A subset B$ and
          $$mu^*(A) = widetilde{mu}(B),$$
          then the statement is true. For example, this is the case if $mu$ is the Lebesgue-measure on $mathbb{R}$ and $mathcal{A}$ is the $sigma$-algebra of all Borel-sets.



          Proof: Take for $A_n$ the corresponding $B_n in mathcal{F}$ with $mu^*(A_n) = widetilde{mu}(B_n)$. Then $x in A$ implies that $x in A_m$ for all $m ge n$, starting at some $n in mathbb{N}$. Thus $x in liminf_{n rightarrow infty} B_n$, because $A_n subset B_n$, and
          $$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$






          share|cite|improve this answer























          • Can you clarify your notation a bit? What are $G_n$? and what is $E$?
            – Bar Alon
            Nov 16 at 8:38










          • Sorry, I have corrected the typos: $G_n$ should be $B_n$.
            – p4sch
            Nov 16 at 8:55













          up vote
          1
          down vote










          up vote
          1
          down vote









          This is only a partial answer. I am not sure, if the statement is in general true!



          If the additive measure $mu$ on $mathcal{A}$ has the property that it extends to a measure $widetilde{mu}$ on a $sigma$-algebra $mathcal{F} supset mathcal{A}$ with the property that for any $A subset X$ there exists a $B in mathcal{F}$ with $A subset B$ and
          $$mu^*(A) = widetilde{mu}(B),$$
          then the statement is true. For example, this is the case if $mu$ is the Lebesgue-measure on $mathbb{R}$ and $mathcal{A}$ is the $sigma$-algebra of all Borel-sets.



          Proof: Take for $A_n$ the corresponding $B_n in mathcal{F}$ with $mu^*(A_n) = widetilde{mu}(B_n)$. Then $x in A$ implies that $x in A_m$ for all $m ge n$, starting at some $n in mathbb{N}$. Thus $x in liminf_{n rightarrow infty} B_n$, because $A_n subset B_n$, and
          $$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$






          share|cite|improve this answer














          This is only a partial answer. I am not sure, if the statement is in general true!



          If the additive measure $mu$ on $mathcal{A}$ has the property that it extends to a measure $widetilde{mu}$ on a $sigma$-algebra $mathcal{F} supset mathcal{A}$ with the property that for any $A subset X$ there exists a $B in mathcal{F}$ with $A subset B$ and
          $$mu^*(A) = widetilde{mu}(B),$$
          then the statement is true. For example, this is the case if $mu$ is the Lebesgue-measure on $mathbb{R}$ and $mathcal{A}$ is the $sigma$-algebra of all Borel-sets.



          Proof: Take for $A_n$ the corresponding $B_n in mathcal{F}$ with $mu^*(A_n) = widetilde{mu}(B_n)$. Then $x in A$ implies that $x in A_m$ for all $m ge n$, starting at some $n in mathbb{N}$. Thus $x in liminf_{n rightarrow infty} B_n$, because $A_n subset B_n$, and
          $$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$







          share|cite|improve this answer














          share|cite|improve this answer



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          edited Nov 16 at 10:42

























          answered Nov 15 at 10:18









          p4sch

          3,630216




          3,630216












          • Can you clarify your notation a bit? What are $G_n$? and what is $E$?
            – Bar Alon
            Nov 16 at 8:38










          • Sorry, I have corrected the typos: $G_n$ should be $B_n$.
            – p4sch
            Nov 16 at 8:55


















          • Can you clarify your notation a bit? What are $G_n$? and what is $E$?
            – Bar Alon
            Nov 16 at 8:38










          • Sorry, I have corrected the typos: $G_n$ should be $B_n$.
            – p4sch
            Nov 16 at 8:55
















          Can you clarify your notation a bit? What are $G_n$? and what is $E$?
          – Bar Alon
          Nov 16 at 8:38




          Can you clarify your notation a bit? What are $G_n$? and what is $E$?
          – Bar Alon
          Nov 16 at 8:38












          Sorry, I have corrected the typos: $G_n$ should be $B_n$.
          – p4sch
          Nov 16 at 8:55




          Sorry, I have corrected the typos: $G_n$ should be $B_n$.
          – p4sch
          Nov 16 at 8:55


















           

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