Unique finite measure on $sigma(mathfrak{A})$ (an algebra of sets)
$begingroup$
I have a statement of a Theorem (possibly some formulation of the Hahn-Kolmogarov theorem?) in my notes as follows:
Let $mathfrak{A}$ be an algebra of subsets of $X$ and let $mu_1,
mu_2$ be finite measures on $sigma(mathfrak{A})$. If $mu_1(A) =
mu_2(A)$ for all $A in mathfrak{A}$, then $mu_1 = mu_2$.
The sketch of the proof I have is this:
Define $mathfrak{M} = {A in sigma(mathfrak{A}) : mu_1(A) =
mu_2(A)}$. Then $mathfrak{M}$ is a monotone collection containing
$mathfrak{A}$. Let $m(mathfrak{A})$ denote the the monotone
collection generated by $mathfrak{A}$. Then $mathfrak{M} supseteq
m(mathfrak{A}) = sigma(mathfrak{A})$.
I’m following all of this.
However, then my notes say:
The result follows from the fact that $mathfrak{M} = sigma(mathfrak{A})$.
I don’t see how that gives the result we need. I would appreciate any clarity on this!
real-analysis analysis measure-theory
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
$endgroup$
|
$begingroup$
I have a statement of a Theorem (possibly some formulation of the Hahn-Kolmogarov theorem?) in my notes as follows:
Let $mathfrak{A}$ be an algebra of subsets of $X$ and let $mu_1,
mu_2$ be finite measures on $sigma(mathfrak{A})$. If $mu_1(A) =
mu_2(A)$ for all $A in mathfrak{A}$, then $mu_1 = mu_2$.
The sketch of the proof I have is this:
Define $mathfrak{M} = {A in sigma(mathfrak{A}) : mu_1(A) =
mu_2(A)}$. Then $mathfrak{M}$ is a monotone collection containing
$mathfrak{A}$. Let $m(mathfrak{A})$ denote the the monotone
collection generated by $mathfrak{A}$. Then $mathfrak{M} supseteq
m(mathfrak{A}) = sigma(mathfrak{A})$.
I’m following all of this.
However, then my notes say:
The result follows from the fact that $mathfrak{M} = sigma(mathfrak{A})$.
I don’t see how that gives the result we need. I would appreciate any clarity on this!
real-analysis analysis measure-theory
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
$endgroup$
1
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40
|
$begingroup$
I have a statement of a Theorem (possibly some formulation of the Hahn-Kolmogarov theorem?) in my notes as follows:
Let $mathfrak{A}$ be an algebra of subsets of $X$ and let $mu_1,
mu_2$ be finite measures on $sigma(mathfrak{A})$. If $mu_1(A) =
mu_2(A)$ for all $A in mathfrak{A}$, then $mu_1 = mu_2$.
The sketch of the proof I have is this:
Define $mathfrak{M} = {A in sigma(mathfrak{A}) : mu_1(A) =
mu_2(A)}$. Then $mathfrak{M}$ is a monotone collection containing
$mathfrak{A}$. Let $m(mathfrak{A})$ denote the the monotone
collection generated by $mathfrak{A}$. Then $mathfrak{M} supseteq
m(mathfrak{A}) = sigma(mathfrak{A})$.
I’m following all of this.
However, then my notes say:
The result follows from the fact that $mathfrak{M} = sigma(mathfrak{A})$.
I don’t see how that gives the result we need. I would appreciate any clarity on this!
real-analysis analysis measure-theory
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
$endgroup$
I have a statement of a Theorem (possibly some formulation of the Hahn-Kolmogarov theorem?) in my notes as follows:
Let $mathfrak{A}$ be an algebra of subsets of $X$ and let $mu_1,
mu_2$ be finite measures on $sigma(mathfrak{A})$. If $mu_1(A) =
mu_2(A)$ for all $A in mathfrak{A}$, then $mu_1 = mu_2$.
The sketch of the proof I have is this:
Define $mathfrak{M} = {A in sigma(mathfrak{A}) : mu_1(A) =
mu_2(A)}$. Then $mathfrak{M}$ is a monotone collection containing
$mathfrak{A}$. Let $m(mathfrak{A})$ denote the the monotone
collection generated by $mathfrak{A}$. Then $mathfrak{M} supseteq
m(mathfrak{A}) = sigma(mathfrak{A})$.
I’m following all of this.
However, then my notes say:
The result follows from the fact that $mathfrak{M} = sigma(mathfrak{A})$.
I don’t see how that gives the result we need. I would appreciate any clarity on this!
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
asked Jan 15 at 16:23
TuringTester69TuringTester69
307213
307213
1
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40
|
1
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40
1
1
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40
|
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
It is a consequence of monotone class theorem: en.wikipedia.org/wiki/Monotone_class_theorem.
$endgroup$
– Song
Jan 15 at 16:40