Stein's real analysis book, chapter 6, Exercise 15
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This is the same question as Infinite product of probability measures is a premeasure but there is one step I don't understand: if $(C_k)_{k=1}^infty$ is a decreasing sequence in $mathcal C$ such that $C_k in mathcal C_k$ for all $k$, and $bigcap_{k=1}^infty C_k = varnothing$, then $m(C_k) to 0$. I don't understand why this is sufficient to prove $sigma$-additivity given finite additivity. I've been stuck on this step for two days, any help is appreciated.
measure-theory
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
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add a comment |
$begingroup$
This is the same question as Infinite product of probability measures is a premeasure but there is one step I don't understand: if $(C_k)_{k=1}^infty$ is a decreasing sequence in $mathcal C$ such that $C_k in mathcal C_k$ for all $k$, and $bigcap_{k=1}^infty C_k = varnothing$, then $m(C_k) to 0$. I don't understand why this is sufficient to prove $sigma$-additivity given finite additivity. I've been stuck on this step for two days, any help is appreciated.
measure-theory
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
$endgroup$
add a comment |
$begingroup$
This is the same question as Infinite product of probability measures is a premeasure but there is one step I don't understand: if $(C_k)_{k=1}^infty$ is a decreasing sequence in $mathcal C$ such that $C_k in mathcal C_k$ for all $k$, and $bigcap_{k=1}^infty C_k = varnothing$, then $m(C_k) to 0$. I don't understand why this is sufficient to prove $sigma$-additivity given finite additivity. I've been stuck on this step for two days, any help is appreciated.
measure-theory
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
$endgroup$
This is the same question as Infinite product of probability measures is a premeasure but there is one step I don't understand: if $(C_k)_{k=1}^infty$ is a decreasing sequence in $mathcal C$ such that $C_k in mathcal C_k$ for all $k$, and $bigcap_{k=1}^infty C_k = varnothing$, then $m(C_k) to 0$. I don't understand why this is sufficient to prove $sigma$-additivity given finite additivity. I've been stuck on this step for two days, any help is appreciated.
measure-theory
measure-theory
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
asked Dec 9 '18 at 20:22
Spiro AgnewSpiro Agnew
132
132
add a comment |
add a comment |
1 Answer
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That's a standard fact about finite measures: If $(E_n)_{n}$ are disjoint and measurable, then $A_n:= bigcup_{k=1}^n E_n$ is increasing and $C_n := A setminus A_n$ decreasing to the emptyset, where $A= bigcup_{n=1}^infty E_n$. Thus $m(C_n) rightarrow 0$, but
$$mu(C_k) = mu(A) - mu(A_n) = mu(A) - sum_{k=1}^n mu(E_k).$$
At this step, we need that $mu$ is finite and finite additive. All in all, we have
$$sum_{k=1}^n mu(E_k) rightarrow mu(A).$$
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
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$begingroup$
That's a standard fact about finite measures: If $(E_n)_{n}$ are disjoint and measurable, then $A_n:= bigcup_{k=1}^n E_n$ is increasing and $C_n := A setminus A_n$ decreasing to the emptyset, where $A= bigcup_{n=1}^infty E_n$. Thus $m(C_n) rightarrow 0$, but
$$mu(C_k) = mu(A) - mu(A_n) = mu(A) - sum_{k=1}^n mu(E_k).$$
At this step, we need that $mu$ is finite and finite additive. All in all, we have
$$sum_{k=1}^n mu(E_k) rightarrow mu(A).$$
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
$endgroup$
add a comment |
$begingroup$
That's a standard fact about finite measures: If $(E_n)_{n}$ are disjoint and measurable, then $A_n:= bigcup_{k=1}^n E_n$ is increasing and $C_n := A setminus A_n$ decreasing to the emptyset, where $A= bigcup_{n=1}^infty E_n$. Thus $m(C_n) rightarrow 0$, but
$$mu(C_k) = mu(A) - mu(A_n) = mu(A) - sum_{k=1}^n mu(E_k).$$
At this step, we need that $mu$ is finite and finite additive. All in all, we have
$$sum_{k=1}^n mu(E_k) rightarrow mu(A).$$
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
$endgroup$
add a comment |
$begingroup$
That's a standard fact about finite measures: If $(E_n)_{n}$ are disjoint and measurable, then $A_n:= bigcup_{k=1}^n E_n$ is increasing and $C_n := A setminus A_n$ decreasing to the emptyset, where $A= bigcup_{n=1}^infty E_n$. Thus $m(C_n) rightarrow 0$, but
$$mu(C_k) = mu(A) - mu(A_n) = mu(A) - sum_{k=1}^n mu(E_k).$$
At this step, we need that $mu$ is finite and finite additive. All in all, we have
$$sum_{k=1}^n mu(E_k) rightarrow mu(A).$$
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
$endgroup$
That's a standard fact about finite measures: If $(E_n)_{n}$ are disjoint and measurable, then $A_n:= bigcup_{k=1}^n E_n$ is increasing and $C_n := A setminus A_n$ decreasing to the emptyset, where $A= bigcup_{n=1}^infty E_n$. Thus $m(C_n) rightarrow 0$, but
$$mu(C_k) = mu(A) - mu(A_n) = mu(A) - sum_{k=1}^n mu(E_k).$$
At this step, we need that $mu$ is finite and finite additive. All in all, we have
$$sum_{k=1}^n mu(E_k) rightarrow mu(A).$$
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
answered Dec 9 '18 at 20:37
p4schp4sch
5,120217
5,120217
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