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0

$begingroup$

I am having a little trouble on solving an exercise from my textbook of Measure Theory.

In the question, $mathbb{S}$ is a semirring and $mathbb{A(S)}$ is the collection of finite disjoint unions of sets of $mathbb{S}$.

Let $mu: mathbb{S}rightarrow mathbb{R}$ a finitely additive measure and
$$mathbb{B(S)}={A subsetOmega |A text{ or } A^c in mathbb{A(S)}}.$$

Show that

i) $mathbb{B(S)}$ is an algebra (that is, it is a ring that contains $Omega$ and therefore contains $B^c, forall Bin mathbb{B(S)}$;

ii) if $mathbb{B(S)}$ is not an algebra then for any given $tin [-infty,+infty]$ there is only one finitely additive measure $mu_t: mathbb{B(S)}rightarrow mathbb{R} cup{t} $ that extends $mu$ and satisfying $mu_t(Omega)=t$.

iii) If $mu$ is sigma-additive and $Omega notin{cup_{n=1}^{infty}S_n|S_n in mathbb{S}}$ then $mu_t$ is sigma-additive.

Here's what I have done so far.

Item i) is pretty straightfoward, just needs some applications of DeMorgan's Laws and the fact that $mathbb{A(S)}$ is a ring.

In item ii), I tried to extend the measure by defining
$$mu_t(A)=t-mu(A^c),$$

if $Anotin mathbb{A(S)}$.

(At this point of the book, we already know how to extend the measure to rings.)

Its unicity is consequence of the fact that, because we want $mu_t$ to be additive:
$$mu_t(Omega)=t=mu_t(A+A^c) = mu_t(A)+mu_t(A^c).$$

My trouble is to prove that $mu_t$ is really finitely additive.

Let $A,B$ disjoint sets of $mathbb{B(S)}$. Since $mathbb{B(S)}$ is an algebra, $(A+B)in mathbb{B(S)}$. Therefore, $(A+B) in mathbb{A(S)}$ or $(A+B)^c in mathbb{A(S)}$. In the first case the additivity ist clear. But I don't know how to proceed in the second case, and any help would be welcome.

Not sure how to proceed in iii) either. Maybe it is necessary to use the definition af $mathbb{A(S)$?

measure-theory

share|cite|improve this question

edited Mar 14 at 16:52

Célio Augusto

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

$endgroup$

add a comment | 

0

$begingroup$

I am having a little trouble on solving an exercise from my textbook of Measure Theory.

In the question, $mathbb{S}$ is a semirring and $mathbb{A(S)}$ is the collection of finite disjoint unions of sets of $mathbb{S}$.

Let $mu: mathbb{S}rightarrow mathbb{R}$ a finitely additive measure and
$$mathbb{B(S)}={A subsetOmega |A text{ or } A^c in mathbb{A(S)}}.$$

Show that

i) $mathbb{B(S)}$ is an algebra (that is, it is a ring that contains $Omega$ and therefore contains $B^c, forall Bin mathbb{B(S)}$;

ii) if $mathbb{B(S)}$ is not an algebra then for any given $tin [-infty,+infty]$ there is only one finitely additive measure $mu_t: mathbb{B(S)}rightarrow mathbb{R} cup{t} $ that extends $mu$ and satisfying $mu_t(Omega)=t$.

iii) If $mu$ is sigma-additive and $Omega notin{cup_{n=1}^{infty}S_n|S_n in mathbb{S}}$ then $mu_t$ is sigma-additive.

Here's what I have done so far.

Item i) is pretty straightfoward, just needs some applications of DeMorgan's Laws and the fact that $mathbb{A(S)}$ is a ring.

In item ii), I tried to extend the measure by defining
$$mu_t(A)=t-mu(A^c),$$

if $Anotin mathbb{A(S)}$.

(At this point of the book, we already know how to extend the measure to rings.)

Its unicity is consequence of the fact that, because we want $mu_t$ to be additive:
$$mu_t(Omega)=t=mu_t(A+A^c) = mu_t(A)+mu_t(A^c).$$

My trouble is to prove that $mu_t$ is really finitely additive.

Let $A,B$ disjoint sets of $mathbb{B(S)}$. Since $mathbb{B(S)}$ is an algebra, $(A+B)in mathbb{B(S)}$. Therefore, $(A+B) in mathbb{A(S)}$ or $(A+B)^c in mathbb{A(S)}$. In the first case the additivity ist clear. But I don't know how to proceed in the second case, and any help would be welcome.

Not sure how to proceed in iii) either. Maybe it is necessary to use the definition af $mathbb{A(S)$?

measure-theory

share|cite|improve this question

edited Mar 14 at 16:52

Célio Augusto

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

$endgroup$

add a comment | 

$begingroup$

I am having a little trouble on solving an exercise from my textbook of Measure Theory.

In the question, $mathbb{S}$ is a semirring and $mathbb{A(S)}$ is the collection of finite disjoint unions of sets of $mathbb{S}$.

Let $mu: mathbb{S}rightarrow mathbb{R}$ a finitely additive measure and
$$mathbb{B(S)}={A subsetOmega |A text{ or } A^c in mathbb{A(S)}}.$$

Show that

i) $mathbb{B(S)}$ is an algebra (that is, it is a ring that contains $Omega$ and therefore contains $B^c, forall Bin mathbb{B(S)}$;

ii) if $mathbb{B(S)}$ is not an algebra then for any given $tin [-infty,+infty]$ there is only one finitely additive measure $mu_t: mathbb{B(S)}rightarrow mathbb{R} cup{t} $ that extends $mu$ and satisfying $mu_t(Omega)=t$.

iii) If $mu$ is sigma-additive and $Omega notin{cup_{n=1}^{infty}S_n|S_n in mathbb{S}}$ then $mu_t$ is sigma-additive.

Here's what I have done so far.

Item i) is pretty straightfoward, just needs some applications of DeMorgan's Laws and the fact that $mathbb{A(S)}$ is a ring.

In item ii), I tried to extend the measure by defining
$$mu_t(A)=t-mu(A^c),$$

if $Anotin mathbb{A(S)}$.

(At this point of the book, we already know how to extend the measure to rings.)

Its unicity is consequence of the fact that, because we want $mu_t$ to be additive:
$$mu_t(Omega)=t=mu_t(A+A^c) = mu_t(A)+mu_t(A^c).$$

My trouble is to prove that $mu_t$ is really finitely additive.

Let $A,B$ disjoint sets of $mathbb{B(S)}$. Since $mathbb{B(S)}$ is an algebra, $(A+B)in mathbb{B(S)}$. Therefore, $(A+B) in mathbb{A(S)}$ or $(A+B)^c in mathbb{A(S)}$. In the first case the additivity ist clear. But I don't know how to proceed in the second case, and any help would be welcome.

Not sure how to proceed in iii) either. Maybe it is necessary to use the definition af $mathbb{A(S)$?

measure-theory

share|cite|improve this question

edited Mar 14 at 16:52

Célio Augusto

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

$endgroup$

share|cite|improve this question

edited Mar 14 at 16:52

Célio Augusto

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

share|cite|improve this question

edited Mar 14 at 16:52

Célio Augusto

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193

asked Mar 14 at 0:31

Célio AugustoCélio Augusto

193