A criterion for checking if a set of measurable functions satisfy a certain property.











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I have recently become interested in analysis, and while checking an old set of notes of mine, I discovered this "theorem" stated without proof in them:



Let $X$ be a compact metric space equipped with the Borel $sigma$-algbera. Let $C_+(X)$ be the set of all the non-negative real-valued continuous functions on $X$. Let $BM_+(X)$ be the set of all the non-negative bounded measurable functions on $X$.




Theorem. Let $Ssubseteq BM_+(X)$ be such that
$1)$ Whenever $f_1, f_2, f_3, ldotsin S$, we have $sum_if_iin S$.
$2)$ Whenever $f, gin S$ with $fgeq g$, we have $f-gin S$.
$3)$ $C_+(X)subseteq S$.

Then $BM_+(X)subseteq S$.




The problem with the above statement is that the first condition on $S$ cannot be satisfied. For if $fin S$ is such that $f(x)>0$ for some $x$, then setting $f_i=f$ for each $igeq 1$ implies by (1) that a non-bounded function is in $F$, contrary to the fact that $Fsubseteq BM_+(X)$.



So the theorem is not true (or rather vacuously true) as stated.



What is the modification that one needs to make in order to make this statement true?



I am interested in this because in some situations it may be easier to see that continuous functions satisfy a certain property. A theorem of the above type will help us conclude that all bounded measurable functions also satisfy that property, provided some reasonable conditions are satisfied.



If you happen to know the correct statement, then can you also please provide a reference?










share|cite|improve this question
























  • Maybe the sum was intended to be finite? (I don't actually know, though.)
    – Clayton
    Oct 4 at 17:40










  • The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
    – MPW
    Oct 4 at 18:11

















up vote
0
down vote

favorite












I have recently become interested in analysis, and while checking an old set of notes of mine, I discovered this "theorem" stated without proof in them:



Let $X$ be a compact metric space equipped with the Borel $sigma$-algbera. Let $C_+(X)$ be the set of all the non-negative real-valued continuous functions on $X$. Let $BM_+(X)$ be the set of all the non-negative bounded measurable functions on $X$.




Theorem. Let $Ssubseteq BM_+(X)$ be such that
$1)$ Whenever $f_1, f_2, f_3, ldotsin S$, we have $sum_if_iin S$.
$2)$ Whenever $f, gin S$ with $fgeq g$, we have $f-gin S$.
$3)$ $C_+(X)subseteq S$.

Then $BM_+(X)subseteq S$.




The problem with the above statement is that the first condition on $S$ cannot be satisfied. For if $fin S$ is such that $f(x)>0$ for some $x$, then setting $f_i=f$ for each $igeq 1$ implies by (1) that a non-bounded function is in $F$, contrary to the fact that $Fsubseteq BM_+(X)$.



So the theorem is not true (or rather vacuously true) as stated.



What is the modification that one needs to make in order to make this statement true?



I am interested in this because in some situations it may be easier to see that continuous functions satisfy a certain property. A theorem of the above type will help us conclude that all bounded measurable functions also satisfy that property, provided some reasonable conditions are satisfied.



If you happen to know the correct statement, then can you also please provide a reference?










share|cite|improve this question
























  • Maybe the sum was intended to be finite? (I don't actually know, though.)
    – Clayton
    Oct 4 at 17:40










  • The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
    – MPW
    Oct 4 at 18:11















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have recently become interested in analysis, and while checking an old set of notes of mine, I discovered this "theorem" stated without proof in them:



Let $X$ be a compact metric space equipped with the Borel $sigma$-algbera. Let $C_+(X)$ be the set of all the non-negative real-valued continuous functions on $X$. Let $BM_+(X)$ be the set of all the non-negative bounded measurable functions on $X$.




Theorem. Let $Ssubseteq BM_+(X)$ be such that
$1)$ Whenever $f_1, f_2, f_3, ldotsin S$, we have $sum_if_iin S$.
$2)$ Whenever $f, gin S$ with $fgeq g$, we have $f-gin S$.
$3)$ $C_+(X)subseteq S$.

Then $BM_+(X)subseteq S$.




The problem with the above statement is that the first condition on $S$ cannot be satisfied. For if $fin S$ is such that $f(x)>0$ for some $x$, then setting $f_i=f$ for each $igeq 1$ implies by (1) that a non-bounded function is in $F$, contrary to the fact that $Fsubseteq BM_+(X)$.



So the theorem is not true (or rather vacuously true) as stated.



What is the modification that one needs to make in order to make this statement true?



I am interested in this because in some situations it may be easier to see that continuous functions satisfy a certain property. A theorem of the above type will help us conclude that all bounded measurable functions also satisfy that property, provided some reasonable conditions are satisfied.



If you happen to know the correct statement, then can you also please provide a reference?










share|cite|improve this question















I have recently become interested in analysis, and while checking an old set of notes of mine, I discovered this "theorem" stated without proof in them:



Let $X$ be a compact metric space equipped with the Borel $sigma$-algbera. Let $C_+(X)$ be the set of all the non-negative real-valued continuous functions on $X$. Let $BM_+(X)$ be the set of all the non-negative bounded measurable functions on $X$.




Theorem. Let $Ssubseteq BM_+(X)$ be such that
$1)$ Whenever $f_1, f_2, f_3, ldotsin S$, we have $sum_if_iin S$.
$2)$ Whenever $f, gin S$ with $fgeq g$, we have $f-gin S$.
$3)$ $C_+(X)subseteq S$.

Then $BM_+(X)subseteq S$.




The problem with the above statement is that the first condition on $S$ cannot be satisfied. For if $fin S$ is such that $f(x)>0$ for some $x$, then setting $f_i=f$ for each $igeq 1$ implies by (1) that a non-bounded function is in $F$, contrary to the fact that $Fsubseteq BM_+(X)$.



So the theorem is not true (or rather vacuously true) as stated.



What is the modification that one needs to make in order to make this statement true?



I am interested in this because in some situations it may be easier to see that continuous functions satisfy a certain property. A theorem of the above type will help us conclude that all bounded measurable functions also satisfy that property, provided some reasonable conditions are satisfied.



If you happen to know the correct statement, then can you also please provide a reference?







real-analysis measure-theory






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edited Oct 4 at 17:52

























asked Oct 4 at 17:29









caffeinemachine

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  • Maybe the sum was intended to be finite? (I don't actually know, though.)
    – Clayton
    Oct 4 at 17:40










  • The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
    – MPW
    Oct 4 at 18:11




















  • Maybe the sum was intended to be finite? (I don't actually know, though.)
    – Clayton
    Oct 4 at 17:40










  • The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
    – MPW
    Oct 4 at 18:11


















Maybe the sum was intended to be finite? (I don't actually know, though.)
– Clayton
Oct 4 at 17:40




Maybe the sum was intended to be finite? (I don't actually know, though.)
– Clayton
Oct 4 at 17:40












The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
– MPW
Oct 4 at 18:11






The intention may be that the $f_i$ are all distinct. So, for example, they could be supported on pairwise disjoint sets. This would not require $S$ to be finite.
– MPW
Oct 4 at 18:11












1 Answer
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0
down vote













$newcommand{set}[1]{{#1}}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$



The correct formulation and the proof is presented as Lemma 2 below.



Let $X$ be a set and $mc D$ be a collection of subsets of $X$.
We say that $mc D$ is a Dynkin system (or $lambda$-system) if



$bullet$ $Xin mc D$.
$bullet$ If $Ain mc D$ then $Xsetminus Ain mc D$.
$bullet$ If $A_1, A_2, A_3, ldots mc D$ such that $A_icap A_j=emptyset$ whenever $ineq j$, then we have $bigcup_{i=1}^infty A_iin mc D$.



Also, we way that a non-empty collection of subsets $mc P$ of $X$ is a $pi$-system if $mc P$ is closed under finite intersections.




Theorem 1.
Dynkin's $pi$-$lambda$ Theorem.
Let $mc P$ be a $pi$-system and $mc D$ be a $lambda$-system on $X$ such that $mc Psubseteq mc D$.
Then $sigma(mc P)subseteq mc D$, where $sigma(mc P)$ denotes the $sigma$-algebra generated by $mc P$.




Let $M_{geq 0}(X)$ denote the set of all the non-negative measurable functions on $X$ and $C_{geq 0}(X)$ denotes the set of all the non-negative continuous function on $X$.




Lemma 2.
Let $Ssubseteq M_{geq 0}(X)$ be such that

1) $C_{geq 0}(X)subseteq S$.

2) If $f, gin F$ with $ggeq f$, then $g-fin S$.

3) If $f_1, f_2, f_3 , ldotsin F$, then we have $sum_{i=1}^infty in S$.

Then $S=M_{geq 0}(X)$.




Proof.
It is enough to show that the indicator functions of Borel sets in $X$ are in $S$.
Let $mc C=set{Ain mc X: chi_Ain S}$.
It is easy to see that $mc C$ is a $lambda$-system.



We show that each open set is in $mc C$.
Let $U$ be an arbitrary open set in $X$.
For each $ngeq 1$, define
$$K_n=set{xin X: d(x, Xsetminus U) geq 1/n}$$
Then each $K_n$ is compact and is contained in $U$.
By Urysohn's lemma, we can find a continuous function $g_n:Xto [0, 1]$ supported in $U$ and which takes the value $1$ on all of $K_n$.
For each $kgeq 1$, define
$$
h_k=maxset{g_1 , ldots, g_k}
$$

Then note that



$bullet$ Each $h_k$ vanishes outside of $U$.
$bullet$ $h_k(x)uparrow 1$ if $xin U$.
$bullet$ $h_1leq h_2leq h_3leq cdots $



Finally define $q_1=h_1$, and $q_i=h_i-h_{i-1}$ for each $igeq 2$.
Then $sum_{i}q_i=chi_U$.
But since each $q_i$ is in $S$, we have that $chi_U$ is also in $S$.
So the collection of all open sets, which is a $pi$-system, is contained in $mc C$.
But then by Theorem 1 we have that $mc C$ contains the $sigma$-algebra generated by the open sets, and we are done.
$blacksquare$






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    $newcommand{set}[1]{{#1}}$
    $newcommand{mc}{mathcal}$
    $newcommand{R}{mathbf R}$



    The correct formulation and the proof is presented as Lemma 2 below.



    Let $X$ be a set and $mc D$ be a collection of subsets of $X$.
    We say that $mc D$ is a Dynkin system (or $lambda$-system) if



    $bullet$ $Xin mc D$.
    $bullet$ If $Ain mc D$ then $Xsetminus Ain mc D$.
    $bullet$ If $A_1, A_2, A_3, ldots mc D$ such that $A_icap A_j=emptyset$ whenever $ineq j$, then we have $bigcup_{i=1}^infty A_iin mc D$.



    Also, we way that a non-empty collection of subsets $mc P$ of $X$ is a $pi$-system if $mc P$ is closed under finite intersections.




    Theorem 1.
    Dynkin's $pi$-$lambda$ Theorem.
    Let $mc P$ be a $pi$-system and $mc D$ be a $lambda$-system on $X$ such that $mc Psubseteq mc D$.
    Then $sigma(mc P)subseteq mc D$, where $sigma(mc P)$ denotes the $sigma$-algebra generated by $mc P$.




    Let $M_{geq 0}(X)$ denote the set of all the non-negative measurable functions on $X$ and $C_{geq 0}(X)$ denotes the set of all the non-negative continuous function on $X$.




    Lemma 2.
    Let $Ssubseteq M_{geq 0}(X)$ be such that

    1) $C_{geq 0}(X)subseteq S$.

    2) If $f, gin F$ with $ggeq f$, then $g-fin S$.

    3) If $f_1, f_2, f_3 , ldotsin F$, then we have $sum_{i=1}^infty in S$.

    Then $S=M_{geq 0}(X)$.




    Proof.
    It is enough to show that the indicator functions of Borel sets in $X$ are in $S$.
    Let $mc C=set{Ain mc X: chi_Ain S}$.
    It is easy to see that $mc C$ is a $lambda$-system.



    We show that each open set is in $mc C$.
    Let $U$ be an arbitrary open set in $X$.
    For each $ngeq 1$, define
    $$K_n=set{xin X: d(x, Xsetminus U) geq 1/n}$$
    Then each $K_n$ is compact and is contained in $U$.
    By Urysohn's lemma, we can find a continuous function $g_n:Xto [0, 1]$ supported in $U$ and which takes the value $1$ on all of $K_n$.
    For each $kgeq 1$, define
    $$
    h_k=maxset{g_1 , ldots, g_k}
    $$

    Then note that



    $bullet$ Each $h_k$ vanishes outside of $U$.
    $bullet$ $h_k(x)uparrow 1$ if $xin U$.
    $bullet$ $h_1leq h_2leq h_3leq cdots $



    Finally define $q_1=h_1$, and $q_i=h_i-h_{i-1}$ for each $igeq 2$.
    Then $sum_{i}q_i=chi_U$.
    But since each $q_i$ is in $S$, we have that $chi_U$ is also in $S$.
    So the collection of all open sets, which is a $pi$-system, is contained in $mc C$.
    But then by Theorem 1 we have that $mc C$ contains the $sigma$-algebra generated by the open sets, and we are done.
    $blacksquare$






    share|cite|improve this answer

























      up vote
      0
      down vote













      $newcommand{set}[1]{{#1}}$
      $newcommand{mc}{mathcal}$
      $newcommand{R}{mathbf R}$



      The correct formulation and the proof is presented as Lemma 2 below.



      Let $X$ be a set and $mc D$ be a collection of subsets of $X$.
      We say that $mc D$ is a Dynkin system (or $lambda$-system) if



      $bullet$ $Xin mc D$.
      $bullet$ If $Ain mc D$ then $Xsetminus Ain mc D$.
      $bullet$ If $A_1, A_2, A_3, ldots mc D$ such that $A_icap A_j=emptyset$ whenever $ineq j$, then we have $bigcup_{i=1}^infty A_iin mc D$.



      Also, we way that a non-empty collection of subsets $mc P$ of $X$ is a $pi$-system if $mc P$ is closed under finite intersections.




      Theorem 1.
      Dynkin's $pi$-$lambda$ Theorem.
      Let $mc P$ be a $pi$-system and $mc D$ be a $lambda$-system on $X$ such that $mc Psubseteq mc D$.
      Then $sigma(mc P)subseteq mc D$, where $sigma(mc P)$ denotes the $sigma$-algebra generated by $mc P$.




      Let $M_{geq 0}(X)$ denote the set of all the non-negative measurable functions on $X$ and $C_{geq 0}(X)$ denotes the set of all the non-negative continuous function on $X$.




      Lemma 2.
      Let $Ssubseteq M_{geq 0}(X)$ be such that

      1) $C_{geq 0}(X)subseteq S$.

      2) If $f, gin F$ with $ggeq f$, then $g-fin S$.

      3) If $f_1, f_2, f_3 , ldotsin F$, then we have $sum_{i=1}^infty in S$.

      Then $S=M_{geq 0}(X)$.




      Proof.
      It is enough to show that the indicator functions of Borel sets in $X$ are in $S$.
      Let $mc C=set{Ain mc X: chi_Ain S}$.
      It is easy to see that $mc C$ is a $lambda$-system.



      We show that each open set is in $mc C$.
      Let $U$ be an arbitrary open set in $X$.
      For each $ngeq 1$, define
      $$K_n=set{xin X: d(x, Xsetminus U) geq 1/n}$$
      Then each $K_n$ is compact and is contained in $U$.
      By Urysohn's lemma, we can find a continuous function $g_n:Xto [0, 1]$ supported in $U$ and which takes the value $1$ on all of $K_n$.
      For each $kgeq 1$, define
      $$
      h_k=maxset{g_1 , ldots, g_k}
      $$

      Then note that



      $bullet$ Each $h_k$ vanishes outside of $U$.
      $bullet$ $h_k(x)uparrow 1$ if $xin U$.
      $bullet$ $h_1leq h_2leq h_3leq cdots $



      Finally define $q_1=h_1$, and $q_i=h_i-h_{i-1}$ for each $igeq 2$.
      Then $sum_{i}q_i=chi_U$.
      But since each $q_i$ is in $S$, we have that $chi_U$ is also in $S$.
      So the collection of all open sets, which is a $pi$-system, is contained in $mc C$.
      But then by Theorem 1 we have that $mc C$ contains the $sigma$-algebra generated by the open sets, and we are done.
      $blacksquare$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        $newcommand{set}[1]{{#1}}$
        $newcommand{mc}{mathcal}$
        $newcommand{R}{mathbf R}$



        The correct formulation and the proof is presented as Lemma 2 below.



        Let $X$ be a set and $mc D$ be a collection of subsets of $X$.
        We say that $mc D$ is a Dynkin system (or $lambda$-system) if



        $bullet$ $Xin mc D$.
        $bullet$ If $Ain mc D$ then $Xsetminus Ain mc D$.
        $bullet$ If $A_1, A_2, A_3, ldots mc D$ such that $A_icap A_j=emptyset$ whenever $ineq j$, then we have $bigcup_{i=1}^infty A_iin mc D$.



        Also, we way that a non-empty collection of subsets $mc P$ of $X$ is a $pi$-system if $mc P$ is closed under finite intersections.




        Theorem 1.
        Dynkin's $pi$-$lambda$ Theorem.
        Let $mc P$ be a $pi$-system and $mc D$ be a $lambda$-system on $X$ such that $mc Psubseteq mc D$.
        Then $sigma(mc P)subseteq mc D$, where $sigma(mc P)$ denotes the $sigma$-algebra generated by $mc P$.




        Let $M_{geq 0}(X)$ denote the set of all the non-negative measurable functions on $X$ and $C_{geq 0}(X)$ denotes the set of all the non-negative continuous function on $X$.




        Lemma 2.
        Let $Ssubseteq M_{geq 0}(X)$ be such that

        1) $C_{geq 0}(X)subseteq S$.

        2) If $f, gin F$ with $ggeq f$, then $g-fin S$.

        3) If $f_1, f_2, f_3 , ldotsin F$, then we have $sum_{i=1}^infty in S$.

        Then $S=M_{geq 0}(X)$.




        Proof.
        It is enough to show that the indicator functions of Borel sets in $X$ are in $S$.
        Let $mc C=set{Ain mc X: chi_Ain S}$.
        It is easy to see that $mc C$ is a $lambda$-system.



        We show that each open set is in $mc C$.
        Let $U$ be an arbitrary open set in $X$.
        For each $ngeq 1$, define
        $$K_n=set{xin X: d(x, Xsetminus U) geq 1/n}$$
        Then each $K_n$ is compact and is contained in $U$.
        By Urysohn's lemma, we can find a continuous function $g_n:Xto [0, 1]$ supported in $U$ and which takes the value $1$ on all of $K_n$.
        For each $kgeq 1$, define
        $$
        h_k=maxset{g_1 , ldots, g_k}
        $$

        Then note that



        $bullet$ Each $h_k$ vanishes outside of $U$.
        $bullet$ $h_k(x)uparrow 1$ if $xin U$.
        $bullet$ $h_1leq h_2leq h_3leq cdots $



        Finally define $q_1=h_1$, and $q_i=h_i-h_{i-1}$ for each $igeq 2$.
        Then $sum_{i}q_i=chi_U$.
        But since each $q_i$ is in $S$, we have that $chi_U$ is also in $S$.
        So the collection of all open sets, which is a $pi$-system, is contained in $mc C$.
        But then by Theorem 1 we have that $mc C$ contains the $sigma$-algebra generated by the open sets, and we are done.
        $blacksquare$






        share|cite|improve this answer












        $newcommand{set}[1]{{#1}}$
        $newcommand{mc}{mathcal}$
        $newcommand{R}{mathbf R}$



        The correct formulation and the proof is presented as Lemma 2 below.



        Let $X$ be a set and $mc D$ be a collection of subsets of $X$.
        We say that $mc D$ is a Dynkin system (or $lambda$-system) if



        $bullet$ $Xin mc D$.
        $bullet$ If $Ain mc D$ then $Xsetminus Ain mc D$.
        $bullet$ If $A_1, A_2, A_3, ldots mc D$ such that $A_icap A_j=emptyset$ whenever $ineq j$, then we have $bigcup_{i=1}^infty A_iin mc D$.



        Also, we way that a non-empty collection of subsets $mc P$ of $X$ is a $pi$-system if $mc P$ is closed under finite intersections.




        Theorem 1.
        Dynkin's $pi$-$lambda$ Theorem.
        Let $mc P$ be a $pi$-system and $mc D$ be a $lambda$-system on $X$ such that $mc Psubseteq mc D$.
        Then $sigma(mc P)subseteq mc D$, where $sigma(mc P)$ denotes the $sigma$-algebra generated by $mc P$.




        Let $M_{geq 0}(X)$ denote the set of all the non-negative measurable functions on $X$ and $C_{geq 0}(X)$ denotes the set of all the non-negative continuous function on $X$.




        Lemma 2.
        Let $Ssubseteq M_{geq 0}(X)$ be such that

        1) $C_{geq 0}(X)subseteq S$.

        2) If $f, gin F$ with $ggeq f$, then $g-fin S$.

        3) If $f_1, f_2, f_3 , ldotsin F$, then we have $sum_{i=1}^infty in S$.

        Then $S=M_{geq 0}(X)$.




        Proof.
        It is enough to show that the indicator functions of Borel sets in $X$ are in $S$.
        Let $mc C=set{Ain mc X: chi_Ain S}$.
        It is easy to see that $mc C$ is a $lambda$-system.



        We show that each open set is in $mc C$.
        Let $U$ be an arbitrary open set in $X$.
        For each $ngeq 1$, define
        $$K_n=set{xin X: d(x, Xsetminus U) geq 1/n}$$
        Then each $K_n$ is compact and is contained in $U$.
        By Urysohn's lemma, we can find a continuous function $g_n:Xto [0, 1]$ supported in $U$ and which takes the value $1$ on all of $K_n$.
        For each $kgeq 1$, define
        $$
        h_k=maxset{g_1 , ldots, g_k}
        $$

        Then note that



        $bullet$ Each $h_k$ vanishes outside of $U$.
        $bullet$ $h_k(x)uparrow 1$ if $xin U$.
        $bullet$ $h_1leq h_2leq h_3leq cdots $



        Finally define $q_1=h_1$, and $q_i=h_i-h_{i-1}$ for each $igeq 2$.
        Then $sum_{i}q_i=chi_U$.
        But since each $q_i$ is in $S$, we have that $chi_U$ is also in $S$.
        So the collection of all open sets, which is a $pi$-system, is contained in $mc C$.
        But then by Theorem 1 we have that $mc C$ contains the $sigma$-algebra generated by the open sets, and we are done.
        $blacksquare$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 8:57









        caffeinemachine

        6,45521250




        6,45521250






























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