Discreet weighted mean inequality












2












$begingroup$



Let ${p_{1}},{p_{2}},ldots,{p_{n}}$ and
${a_{1}},{a_{2}},ldots,{a_{n}}$ be positive real numbers and let
$r$ be a real number. Then for $rne0$ , we define
${M_{r}}(a,p)={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}}$
and for $r=0$ , we define
${M_{0}}(a,p)={left({a_{1}^{{p_{1}}}a_{2}^{{p_{2}}}cdots
a_{n}^{{p_{n}}}}right)}^{1/sumnolimits _{i=1}^{n}p_i}$ . Then prove that $
{M_{{k_{1}}}}(a,p)geqslant{M_{{k_{2}}}}(a,p) $ if $k_{1}geqslant
k_{2}$.




How to prove this generalized theorem? I have found this in a book without any proof. So can anyone show me?






calculus real-analysis inequality means holder-inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
    $endgroup$
    – MathematicianByMistake
    May 9 '16 at 7:18








  • 1




    $begingroup$
    Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
    $endgroup$
    – Aritra Das
    May 9 '16 at 7:18










  • $begingroup$
    This is more than the generalized AM GM. It includes all weighted means.
    $endgroup$
    – Perth
    May 9 '16 at 7:41


















2












$begingroup$



Let ${p_{1}},{p_{2}},ldots,{p_{n}}$ and
${a_{1}},{a_{2}},ldots,{a_{n}}$ be positive real numbers and let
$r$ be a real number. Then for $rne0$ , we define
${M_{r}}(a,p)={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}}$
and for $r=0$ , we define
${M_{0}}(a,p)={left({a_{1}^{{p_{1}}}a_{2}^{{p_{2}}}cdots
a_{n}^{{p_{n}}}}right)}^{1/sumnolimits _{i=1}^{n}p_i}$ . Then prove that $
{M_{{k_{1}}}}(a,p)geqslant{M_{{k_{2}}}}(a,p) $ if $k_{1}geqslant
k_{2}$.




How to prove this generalized theorem? I have found this in a book without any proof. So can anyone show me?






calculus real-analysis inequality means holder-inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
    $endgroup$
    – MathematicianByMistake
    May 9 '16 at 7:18








  • 1




    $begingroup$
    Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
    $endgroup$
    – Aritra Das
    May 9 '16 at 7:18










  • $begingroup$
    This is more than the generalized AM GM. It includes all weighted means.
    $endgroup$
    – Perth
    May 9 '16 at 7:41
















2












2








2


1



$begingroup$



Let ${p_{1}},{p_{2}},ldots,{p_{n}}$ and
${a_{1}},{a_{2}},ldots,{a_{n}}$ be positive real numbers and let
$r$ be a real number. Then for $rne0$ , we define
${M_{r}}(a,p)={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}}$
and for $r=0$ , we define
${M_{0}}(a,p)={left({a_{1}^{{p_{1}}}a_{2}^{{p_{2}}}cdots
a_{n}^{{p_{n}}}}right)}^{1/sumnolimits _{i=1}^{n}p_i}$ . Then prove that $
{M_{{k_{1}}}}(a,p)geqslant{M_{{k_{2}}}}(a,p) $ if $k_{1}geqslant
k_{2}$.




How to prove this generalized theorem? I have found this in a book without any proof. So can anyone show me?






calculus real-analysis inequality means holder-inequality







share|cite|improve this question











$endgroup$





Let ${p_{1}},{p_{2}},ldots,{p_{n}}$ and
${a_{1}},{a_{2}},ldots,{a_{n}}$ be positive real numbers and let
$r$ be a real number. Then for $rne0$ , we define
${M_{r}}(a,p)={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}}$
and for $r=0$ , we define
${M_{0}}(a,p)={left({a_{1}^{{p_{1}}}a_{2}^{{p_{2}}}cdots
a_{n}^{{p_{n}}}}right)}^{1/sumnolimits _{i=1}^{n}p_i}$ . Then prove that $
{M_{{k_{1}}}}(a,p)geqslant{M_{{k_{2}}}}(a,p) $ if $k_{1}geqslant
k_{2}$.




How to prove this generalized theorem? I have found this in a book without any proof. So can anyone show me?






calculus real-analysis inequality means holder-inequality




calculus real-analysis inequality means holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '17 at 22:32









Guy Fsone

17.1k42873




17.1k42873










asked May 9 '16 at 7:16









PerthPerth

906




906








  • 1




    $begingroup$
    Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
    $endgroup$
    – MathematicianByMistake
    May 9 '16 at 7:18








  • 1




    $begingroup$
    Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
    $endgroup$
    – Aritra Das
    May 9 '16 at 7:18










  • $begingroup$
    This is more than the generalized AM GM. It includes all weighted means.
    $endgroup$
    – Perth
    May 9 '16 at 7:41
















  • 1




    $begingroup$
    Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
    $endgroup$
    – MathematicianByMistake
    May 9 '16 at 7:18








  • 1




    $begingroup$
    Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
    $endgroup$
    – Aritra Das
    May 9 '16 at 7:18










  • $begingroup$
    This is more than the generalized AM GM. It includes all weighted means.
    $endgroup$
    – Perth
    May 9 '16 at 7:41










1




1




$begingroup$
Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
$endgroup$
– MathematicianByMistake
May 9 '16 at 7:18






$begingroup$
Check here: math.stackexchange.com/questions/922401/… A proof is presened by the OP.
$endgroup$
– MathematicianByMistake
May 9 '16 at 7:18






1




1




$begingroup$
Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
$endgroup$
– Aritra Das
May 9 '16 at 7:18




$begingroup$
Jensen's inequality comes to mind. Have a look : en.wikipedia.org/wiki/Jensen%27s_inequality
$endgroup$
– Aritra Das
May 9 '16 at 7:18












$begingroup$
This is more than the generalized AM GM. It includes all weighted means.
$endgroup$
– Perth
May 9 '16 at 7:41






$begingroup$
This is more than the generalized AM GM. It includes all weighted means.
$endgroup$
– Perth
May 9 '16 at 7:41












2 Answers
2






active

oldest

votes


















0












$begingroup$

In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :



(i) Bernoulli inequality,



(ii) the weighted AM-GM inequality,



(iii) Holder inequality,



(iv) the weighted power mean inequality,



(v) Minkovski inequality,



(vi) Radon inequality.



See the theorem 2.3 for that .






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
    $endgroup$
    – Guy Fsone
    Dec 19 '17 at 18:22



















0












$begingroup$

Let $$w_{i} = frac{p_{i}}{sum_{i=1}^np_{i}}impliessum_{i=1}^nw_{i}=1 $$




we recall that by Holder inequality one has
$$sum_{i=1}^nw_{i}A_i B_i
le left(sum_{i=1}^nw_{i}A_i^{color{red}{q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}B_i^{color{red}{q'}}right)^{color{red}{1/q'}}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1 $$ with $qge1$




Therefore, for $kge r$ if we let $q=frac{k}{r}ge 1$ we have, $$color{blue}{M_r(a,p)}={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}} =left(sum_{i=1}^nw_{i}a_{i}^{r}right)^{1/r}
\=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{r/k}}right)^{1/r}=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{1/q}}cdotcolor{blue}{1}right)^{1/r}
\overset{Holder}{le}left(left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{q/q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}1^{color{red}{1/q'}}right)^{color{red}{q'/q}}right)^{1/r}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1
\=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/qr}=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/k}=color{blue}{M_k(a,p)}~~~~~~qr=k $$
that is $$M_{k}(a,p)geqslant M_{r}(a,p)~~~~~kge r$$
where






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :



    (i) Bernoulli inequality,



    (ii) the weighted AM-GM inequality,



    (iii) Holder inequality,



    (iv) the weighted power mean inequality,



    (v) Minkovski inequality,



    (vi) Radon inequality.



    See the theorem 2.3 for that .






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
      $endgroup$
      – Guy Fsone
      Dec 19 '17 at 18:22
















    0












    $begingroup$

    In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :



    (i) Bernoulli inequality,



    (ii) the weighted AM-GM inequality,



    (iii) Holder inequality,



    (iv) the weighted power mean inequality,



    (v) Minkovski inequality,



    (vi) Radon inequality.



    See the theorem 2.3 for that .






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
      $endgroup$
      – Guy Fsone
      Dec 19 '17 at 18:22














    0












    0








    0





    $begingroup$

    In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :



    (i) Bernoulli inequality,



    (ii) the weighted AM-GM inequality,



    (iii) Holder inequality,



    (iv) the weighted power mean inequality,



    (v) Minkovski inequality,



    (vi) Radon inequality.



    See the theorem 2.3 for that .






    share|cite|improve this answer









    $endgroup$



    In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :



    (i) Bernoulli inequality,



    (ii) the weighted AM-GM inequality,



    (iii) Holder inequality,



    (iv) the weighted power mean inequality,



    (v) Minkovski inequality,



    (vi) Radon inequality.



    See the theorem 2.3 for that .







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 '17 at 17:53







    user448747



















    • $begingroup$
      why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
      $endgroup$
      – Guy Fsone
      Dec 19 '17 at 18:22


















    • $begingroup$
      why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
      $endgroup$
      – Guy Fsone
      Dec 19 '17 at 18:22
















    $begingroup$
    why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
    $endgroup$
    – Guy Fsone
    Dec 19 '17 at 18:22




    $begingroup$
    why do you need all that? Holder inequality or Jessen inequality is sufficiently enough
    $endgroup$
    – Guy Fsone
    Dec 19 '17 at 18:22











    0












    $begingroup$

    Let $$w_{i} = frac{p_{i}}{sum_{i=1}^np_{i}}impliessum_{i=1}^nw_{i}=1 $$




    we recall that by Holder inequality one has
    $$sum_{i=1}^nw_{i}A_i B_i
    le left(sum_{i=1}^nw_{i}A_i^{color{red}{q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}B_i^{color{red}{q'}}right)^{color{red}{1/q'}}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1 $$ with $qge1$




    Therefore, for $kge r$ if we let $q=frac{k}{r}ge 1$ we have, $$color{blue}{M_r(a,p)}={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}} =left(sum_{i=1}^nw_{i}a_{i}^{r}right)^{1/r}
    \=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{r/k}}right)^{1/r}=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{1/q}}cdotcolor{blue}{1}right)^{1/r}
    \overset{Holder}{le}left(left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{q/q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}1^{color{red}{1/q'}}right)^{color{red}{q'/q}}right)^{1/r}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1
    \=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/qr}=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/k}=color{blue}{M_k(a,p)}~~~~~~qr=k $$
    that is $$M_{k}(a,p)geqslant M_{r}(a,p)~~~~~kge r$$
    where






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $$w_{i} = frac{p_{i}}{sum_{i=1}^np_{i}}impliessum_{i=1}^nw_{i}=1 $$




      we recall that by Holder inequality one has
      $$sum_{i=1}^nw_{i}A_i B_i
      le left(sum_{i=1}^nw_{i}A_i^{color{red}{q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}B_i^{color{red}{q'}}right)^{color{red}{1/q'}}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1 $$ with $qge1$




      Therefore, for $kge r$ if we let $q=frac{k}{r}ge 1$ we have, $$color{blue}{M_r(a,p)}={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}} =left(sum_{i=1}^nw_{i}a_{i}^{r}right)^{1/r}
      \=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{r/k}}right)^{1/r}=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{1/q}}cdotcolor{blue}{1}right)^{1/r}
      \overset{Holder}{le}left(left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{q/q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}1^{color{red}{1/q'}}right)^{color{red}{q'/q}}right)^{1/r}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1
      \=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/qr}=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/k}=color{blue}{M_k(a,p)}~~~~~~qr=k $$
      that is $$M_{k}(a,p)geqslant M_{r}(a,p)~~~~~kge r$$
      where






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $$w_{i} = frac{p_{i}}{sum_{i=1}^np_{i}}impliessum_{i=1}^nw_{i}=1 $$




        we recall that by Holder inequality one has
        $$sum_{i=1}^nw_{i}A_i B_i
        le left(sum_{i=1}^nw_{i}A_i^{color{red}{q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}B_i^{color{red}{q'}}right)^{color{red}{1/q'}}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1 $$ with $qge1$




        Therefore, for $kge r$ if we let $q=frac{k}{r}ge 1$ we have, $$color{blue}{M_r(a,p)}={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}} =left(sum_{i=1}^nw_{i}a_{i}^{r}right)^{1/r}
        \=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{r/k}}right)^{1/r}=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{1/q}}cdotcolor{blue}{1}right)^{1/r}
        \overset{Holder}{le}left(left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{q/q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}1^{color{red}{1/q'}}right)^{color{red}{q'/q}}right)^{1/r}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1
        \=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/qr}=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/k}=color{blue}{M_k(a,p)}~~~~~~qr=k $$
        that is $$M_{k}(a,p)geqslant M_{r}(a,p)~~~~~kge r$$
        where






        share|cite|improve this answer











        $endgroup$



        Let $$w_{i} = frac{p_{i}}{sum_{i=1}^np_{i}}impliessum_{i=1}^nw_{i}=1 $$




        we recall that by Holder inequality one has
        $$sum_{i=1}^nw_{i}A_i B_i
        le left(sum_{i=1}^nw_{i}A_i^{color{red}{q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}B_i^{color{red}{q'}}right)^{color{red}{1/q'}}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1 $$ with $qge1$




        Therefore, for $kge r$ if we let $q=frac{k}{r}ge 1$ we have, $$color{blue}{M_r(a,p)}={left({frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+cdots+{p_{n}}}}}right)^{1/r}} =left(sum_{i=1}^nw_{i}a_{i}^{r}right)^{1/r}
        \=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{r/k}}right)^{1/r}=left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{1/q}}cdotcolor{blue}{1}right)^{1/r}
        \overset{Holder}{le}left(left(sum_{i=1}^nw_{i}left(a_{i}^{k}right)^{color{red}{q/q}}right)^{color{red}{1/q}}left(sum_{i=1}^nw_{i}1^{color{red}{1/q'}}right)^{color{red}{q'/q}}right)^{1/r}~~~~~~{color{red}{1/q}}+{color{red}{1/q'}}=1
        \=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/qr}=left(sum_{i=1}^nw_{i}a_{i}^{k}right)^{1/k}=color{blue}{M_k(a,p)}~~~~~~qr=k $$
        that is $$M_{k}(a,p)geqslant M_{r}(a,p)~~~~~kge r$$
        where







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '17 at 18:32

























        answered Dec 19 '17 at 18:20









        Guy FsoneGuy Fsone

        17.1k42873




        17.1k42873






























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            Mario Kart Wii

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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            Partial Derivative Guidance.

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            0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
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            Understanding the size os this class of aleatory events

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            0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
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