Iterated induced outermeasures from premeasure












1












$begingroup$


Let $X$ be a set, $mathcal{A} subseteq mathcal{P}(X)$ be an algebra (i.e., a set closed under complements and finite unions), $mu_0$ be a premeasure defined on $mathcal{A}$, and $mathcal{M}$ be the $sigma$-algebra generated by $mathcal{A}$. Denote $mu^*$ to be the outer measure induced by $mu_0$; explicitly, for $E in mathcal{P}(X)$, define
$$mu^*(E) = inf left{sum_{i=1}^{infty}mu_0(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{A} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
Further, define
$$mathcal{M}^*=left{E in mathcal{P}(X): forall K in mathcal{P}(X), ;mu^*(K) = mu^*(Kcap E) + mu^*(Kcap E^c)right}.$$
That $mathcal{M}^*$ is a $sigma$-algebra, $mathcal{A} subseteq mathcal{M}^*$, and hence $mathcal{M} subseteq mathcal{M}^*$ are known facts. When does $mathcal{M} = mathcal{M}^*$? Since $mu^* Big|_{mathcal{A}} = mu_0$, $mu^*Big|_{mathcal{M}}$ is a measure that extends $mu_0$, and $mu^* Big|_{mathcal{M}^*} $ is a complete measure, the above equality would imply $mu = mu^*Big|_{mathcal{M}} = mu^*Big|_{mathcal{M}^*}$ is a complete measure extending $mu_0$.



Now, leaving the question of the above equality aside, denote $mu^{**}$ to be the outer measure induced by $mu = mu^*Big|_{mathcal{M}}$; that is,
for $E in mathcal{P}(X)$, define
$$mu^{**}(E) = inf left{sum_{i=1}^{infty}mu(E_i) = sum_{i=1}^{infty}mu^*(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{M} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
What is the relationship between $mu^*$ and $mu^{**}$? More generally, if I define $mu^{overbrace{*ldots*}^{n+1}}$ to be the outer measure induced by $mu^{overbrace{*ldots*}^{n}}Big|_{mathcal{M}}$ inductively for $n in mathbb{N}$ as above, what is the relationship between these outer measures? What is the pointwise limit of this sequence of induced outer measures?



Examples elucidating your arguments will be much appreciated.






measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
    $endgroup$
    – Masacroso
    Dec 12 '18 at 1:55


















1












$begingroup$


Let $X$ be a set, $mathcal{A} subseteq mathcal{P}(X)$ be an algebra (i.e., a set closed under complements and finite unions), $mu_0$ be a premeasure defined on $mathcal{A}$, and $mathcal{M}$ be the $sigma$-algebra generated by $mathcal{A}$. Denote $mu^*$ to be the outer measure induced by $mu_0$; explicitly, for $E in mathcal{P}(X)$, define
$$mu^*(E) = inf left{sum_{i=1}^{infty}mu_0(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{A} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
Further, define
$$mathcal{M}^*=left{E in mathcal{P}(X): forall K in mathcal{P}(X), ;mu^*(K) = mu^*(Kcap E) + mu^*(Kcap E^c)right}.$$
That $mathcal{M}^*$ is a $sigma$-algebra, $mathcal{A} subseteq mathcal{M}^*$, and hence $mathcal{M} subseteq mathcal{M}^*$ are known facts. When does $mathcal{M} = mathcal{M}^*$? Since $mu^* Big|_{mathcal{A}} = mu_0$, $mu^*Big|_{mathcal{M}}$ is a measure that extends $mu_0$, and $mu^* Big|_{mathcal{M}^*} $ is a complete measure, the above equality would imply $mu = mu^*Big|_{mathcal{M}} = mu^*Big|_{mathcal{M}^*}$ is a complete measure extending $mu_0$.



Now, leaving the question of the above equality aside, denote $mu^{**}$ to be the outer measure induced by $mu = mu^*Big|_{mathcal{M}}$; that is,
for $E in mathcal{P}(X)$, define
$$mu^{**}(E) = inf left{sum_{i=1}^{infty}mu(E_i) = sum_{i=1}^{infty}mu^*(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{M} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
What is the relationship between $mu^*$ and $mu^{**}$? More generally, if I define $mu^{overbrace{*ldots*}^{n+1}}$ to be the outer measure induced by $mu^{overbrace{*ldots*}^{n}}Big|_{mathcal{M}}$ inductively for $n in mathbb{N}$ as above, what is the relationship between these outer measures? What is the pointwise limit of this sequence of induced outer measures?



Examples elucidating your arguments will be much appreciated.






measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
    $endgroup$
    – Masacroso
    Dec 12 '18 at 1:55
















1












1








1





$begingroup$


Let $X$ be a set, $mathcal{A} subseteq mathcal{P}(X)$ be an algebra (i.e., a set closed under complements and finite unions), $mu_0$ be a premeasure defined on $mathcal{A}$, and $mathcal{M}$ be the $sigma$-algebra generated by $mathcal{A}$. Denote $mu^*$ to be the outer measure induced by $mu_0$; explicitly, for $E in mathcal{P}(X)$, define
$$mu^*(E) = inf left{sum_{i=1}^{infty}mu_0(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{A} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
Further, define
$$mathcal{M}^*=left{E in mathcal{P}(X): forall K in mathcal{P}(X), ;mu^*(K) = mu^*(Kcap E) + mu^*(Kcap E^c)right}.$$
That $mathcal{M}^*$ is a $sigma$-algebra, $mathcal{A} subseteq mathcal{M}^*$, and hence $mathcal{M} subseteq mathcal{M}^*$ are known facts. When does $mathcal{M} = mathcal{M}^*$? Since $mu^* Big|_{mathcal{A}} = mu_0$, $mu^*Big|_{mathcal{M}}$ is a measure that extends $mu_0$, and $mu^* Big|_{mathcal{M}^*} $ is a complete measure, the above equality would imply $mu = mu^*Big|_{mathcal{M}} = mu^*Big|_{mathcal{M}^*}$ is a complete measure extending $mu_0$.



Now, leaving the question of the above equality aside, denote $mu^{**}$ to be the outer measure induced by $mu = mu^*Big|_{mathcal{M}}$; that is,
for $E in mathcal{P}(X)$, define
$$mu^{**}(E) = inf left{sum_{i=1}^{infty}mu(E_i) = sum_{i=1}^{infty}mu^*(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{M} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
What is the relationship between $mu^*$ and $mu^{**}$? More generally, if I define $mu^{overbrace{*ldots*}^{n+1}}$ to be the outer measure induced by $mu^{overbrace{*ldots*}^{n}}Big|_{mathcal{M}}$ inductively for $n in mathbb{N}$ as above, what is the relationship between these outer measures? What is the pointwise limit of this sequence of induced outer measures?



Examples elucidating your arguments will be much appreciated.






measure-theory







share|cite|improve this question











$endgroup$




Let $X$ be a set, $mathcal{A} subseteq mathcal{P}(X)$ be an algebra (i.e., a set closed under complements and finite unions), $mu_0$ be a premeasure defined on $mathcal{A}$, and $mathcal{M}$ be the $sigma$-algebra generated by $mathcal{A}$. Denote $mu^*$ to be the outer measure induced by $mu_0$; explicitly, for $E in mathcal{P}(X)$, define
$$mu^*(E) = inf left{sum_{i=1}^{infty}mu_0(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{A} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
Further, define
$$mathcal{M}^*=left{E in mathcal{P}(X): forall K in mathcal{P}(X), ;mu^*(K) = mu^*(Kcap E) + mu^*(Kcap E^c)right}.$$
That $mathcal{M}^*$ is a $sigma$-algebra, $mathcal{A} subseteq mathcal{M}^*$, and hence $mathcal{M} subseteq mathcal{M}^*$ are known facts. When does $mathcal{M} = mathcal{M}^*$? Since $mu^* Big|_{mathcal{A}} = mu_0$, $mu^*Big|_{mathcal{M}}$ is a measure that extends $mu_0$, and $mu^* Big|_{mathcal{M}^*} $ is a complete measure, the above equality would imply $mu = mu^*Big|_{mathcal{M}} = mu^*Big|_{mathcal{M}^*}$ is a complete measure extending $mu_0$.



Now, leaving the question of the above equality aside, denote $mu^{**}$ to be the outer measure induced by $mu = mu^*Big|_{mathcal{M}}$; that is,
for $E in mathcal{P}(X)$, define
$$mu^{**}(E) = inf left{sum_{i=1}^{infty}mu(E_i) = sum_{i=1}^{infty}mu^*(E_i): {E_i}_{i=1}^{infty}subseteq mathcal{M} text{ is disjoint and } E subseteq bigcup_{i=1}^{infty}E_iright}.$$
What is the relationship between $mu^*$ and $mu^{**}$? More generally, if I define $mu^{overbrace{*ldots*}^{n+1}}$ to be the outer measure induced by $mu^{overbrace{*ldots*}^{n}}Big|_{mathcal{M}}$ inductively for $n in mathbb{N}$ as above, what is the relationship between these outer measures? What is the pointwise limit of this sequence of induced outer measures?



Examples elucidating your arguments will be much appreciated.






measure-theory




measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 2:06







SystematicDisintegration

















asked Dec 12 '18 at 1:46









SystematicDisintegrationSystematicDisintegration

279211




279211








  • 1




    $begingroup$
    the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
    $endgroup$
    – Masacroso
    Dec 12 '18 at 1:55
















  • 1




    $begingroup$
    the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
    $endgroup$
    – Masacroso
    Dec 12 '18 at 1:55










1




1




$begingroup$
the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
$endgroup$
– Masacroso
Dec 12 '18 at 1:55






$begingroup$
the condition of ${E_i}_{iinBbb N}$ be disjoint is suprefluous, because they are elements of a $sigma$-algebra, that is, for every ${B_i}_{iinBbb N}$ of non-disjoint sets we can build a sequence of pair-wise disjoint sets ${E_i}$, by example $E_i:=B_isetminus(bigcup_{j=0}^{i-1} B_j)$, such that $bigcup_{iinBbb N} B_i=bigcup_{iinBbb N} E_i$
$endgroup$
– Masacroso
Dec 12 '18 at 1:55












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