Measure preserving

Shadow · **Posted:** Sat, 28 Jul 2012 22:50:27 UTC

If I'm trying to show some action is measure preserving on a Borel $\sigma$ -algebra, is it enough to show it on the basic open sets? Or, more generally, is it enough to show it on any system of generators?

If not, then what if I add any combination of the assumptions of regularity or finite measure? I feel like that *should* be closer to reasonable as the measure is continuous then, and I should only have to show it plays well with finite intersections and arbitrary unions of basic open sets since I can then take the inf over open sets containing my set.

Any help appreciated.

outermeasure · **Posted:** Sun, 29 Jul 2012 02:22:31 UTC

Shadow wrote:

If I'm trying to show some action is measure preserving on a Borel $\sigma$ -algebra, is it enough to show it on the basic open sets? Or, more generally, is it enough to show it on any system of generators?

If not, then what if I add any combination of the assumptions of regularity or finite measure? I feel like that *should* be closer to reasonable as the measure is continuous then, and I should only have to show it plays well with finite intersections and arbitrary unions of basic open sets since I can then take the inf over open sets containing my set.

Any help appreciated.

Depends on what you mean by generators. For example, $\mathcal{X}=\{\text{o}\text{pen sets not containing }0\}$ generates the Borel $\sigma$ -algebra on $\mathbb{R}$ (i.e. $\sigma(\mathcal{X})=\mathcal{B}(\mathbb{R})$ ), but the zero measure and the Dirac delta $\delta_0$ agree on $\mathcal{X}$ .

If the measure is $\sigma$ -finite, then by the uniqueness of extension (Caratheodory/Hahn-Kolmogorov), it is enough to show it on any $\pi$ -system generating $\mathcal{B}$ , e.g. all open intervals in $\mathbb{R}$ .

Shadow · **Posted:** Mon, 30 Jul 2012 00:20:29 UTC

I see, that's because $\sigma$ -finiteness allows us to break up the sets into disjoint pieces we can add up, and continuity of the measure?

Still, I'm not quite sure why a $\pi$ system is enough, shouldn't there be closure under (relative) compliments as well? After all, {open sets not containing 0} are a $\pi$ -system, are they not? It seems like you're appealing to the Caratheodory theorem, but that needs the compliment bit as well, or I'm misunderstanding where you're coming from on that.

outermeasure · **Posted:** Mon, 30 Jul 2012 03:48:02 UTC

Shadow wrote:

I see, that's because $\sigma$ -finiteness allows us to break up the sets into disjoint pieces we can add up, and continuity of the measure?

Still, I'm not quite sure why a $\pi$ system is enough, shouldn't there be closure under (relative) compliments as well? After all, {open sets not containing 0} are a $\pi$ -system, are they not? It seems like you're appealing to the Caratheodory theorem, but that needs the compliment bit as well, or I'm misunderstanding where you're coming from on that.

Oops, I don't mean $\sigma$ -finiteness on $(X,\mathcal{B})$ , but $\sigma$ -finite using sets in the pi-system $\mathcal{Q}$ , i.e. there exists $X_i\in\mathcal{Q}$ such that $\bigcup X_i=X$ (that disqualifies the {open sets not containing 0} example because you can never cover 0).

$\sigma$ -finiteness allows us to add up and subtract, so we can get relative complements:
$\mu(X_i-X_j)=\mu(X_i)-\mu(X_i\cap X_j)$ etc (so you can enlarge $\mathcal{Q}$ by assuming X_i

are nested) and
$\mu(A-B)=\lim_i(\mu(A\cap X_i)-\mu(B\cap X_i))$ (for $B\subset A$ ).
Alternatively, we can also appeal to the uniqueness of $\mu\mathrel{\raisebox{.6ex}{\ensuremath{\llcorner}}\mkern-9.4mu-} X_i$ (if two finite measures agree on a pi-system and have the same total mass, then Dynkin => they agree on the sigma-algebra generated) and take limit

Shadow · **Posted:** Mon, 30 Jul 2012 03:54:39 UTC

outermeasure wrote:

Shadow wrote:

I see, that's because $\sigma$ -finiteness allows us to break up the sets into disjoint pieces we can add up, and continuity of the measure?

Still, I'm not quite sure why a $\pi$ system is enough, shouldn't there be closure under (relative) compliments as well? After all, {open sets not containing 0} are a $\pi$ -system, are they not? It seems like you're appealing to the Caratheodory theorem, but that needs the compliment bit as well, or I'm misunderstanding where you're coming from on that.

Oops, I don't mean $\sigma$ -finiteness on $(X,\mathcal{B})$ , but $\sigma$ -finite using sets in the pi-system $\mathcal{Q}$ , i.e. there exists $X_i\in\mathcal{Q}$ such that $\bigcup X_i=X$ (that disqualifies the {open sets not containing 0} example because you can never cover 0).

$\sigma$ -finiteness allows us to add up and subtract, so we can get relative complements:
$\mu(X_i-X_j)=\mu(X_i)-\mu(X_i\cap X_j)$ etc (so you can enlarge $\mathcal{Q}$ by assuming X_i

are nested) and
$\mu(A-B)=\lim_i(\mu(A\cap X_i)-\mu(B\cap X_i))$ (for $B\subset A$ ).

I'm just a little confused though, what does $\sigma$ -finiteness have to do with whether or not the $\pi$ -system covers the space? I get what you meant to say now, this is just a question on terminology.

outermeasure · **Posted:** Mon, 30 Jul 2012 04:14:30 UTC

Shadow wrote:

outermeasure wrote:

Shadow wrote:

I see, that's because $\sigma$ -finiteness allows us to break up the sets into disjoint pieces we can add up, and continuity of the measure?

Still, I'm not quite sure why a $\pi$ system is enough, shouldn't there be closure under (relative) compliments as well? After all, {open sets not containing 0} are a $\pi$ -system, are they not? It seems like you're appealing to the Caratheodory theorem, but that needs the compliment bit as well, or I'm misunderstanding where you're coming from on that.

Oops, I don't mean $\sigma$ -finiteness on $(X,\mathcal{B})$ , but $\sigma$ -finite using sets in the pi-system $\mathcal{Q}$ , i.e. there exists $X_i\in\mathcal{Q}$ such that $\bigcup X_i=X$ (that disqualifies the {open sets not containing 0} example because you can never cover 0).

$\sigma$ -finiteness allows us to add up and subtract, so we can get relative complements:
$\mu(X_i-X_j)=\mu(X_i)-\mu(X_i\cap X_j)$ etc (so you can enlarge $\mathcal{Q}$ by assuming X_i

are nested) and
$\mu(A-B)=\lim_i(\mu(A\cap X_i)-\mu(B\cap X_i))$ (for $B\subset A$ ).

I'm just a little confused though, what does $\sigma$ -finiteness have to do with whether or not the $\pi$ -system covers the space? I get what you meant to say now, this is just a question on terminology.

It has nothing to do there, except I was thinking about the general statement of uniqueness of measure extended from pi-system to its sigma-algebra:

Theorem: Let $\lambda ,\mu$ be measures on a measurable space $(X,\mathcal{A})$ . Suppose that

is a $\pi$ -system on

generating $\mathcal{A}$ such that $\lambda=\mu$ on

and that there exists a sequence $S_n\in A$ with $\bigcup_{n=1}^\infty S_n=X$ and $\lambda(S_n)<\infty$ . Then $\lambda=\mu$ .
Proof: Apply Dynkin's $\pi\text{-}\lambda$ theorem to $\{S\in\mathcal{A}\mid\lambda(S)=\mu(S)\}$ . Details omitted.

The non-uniqueness when you don't have S_n

is demonstrated by $X=\mathbb{Q}$ , $A=\{[a,b)\cap\mathbb{Q}\mid a,b\in\mathbb{Q}\}$ , when any multiple of counting measure will give the same result (infinity on nonempty) on

.

Shadow · **Posted:** Mon, 30 Jul 2012 04:23:35 UTC

outermeasure wrote:

Shadow wrote:

outermeasure wrote:

Shadow wrote:

I see, that's because $\sigma$ -finiteness allows us to break up the sets into disjoint pieces we can add up, and continuity of the measure?

Still, I'm not quite sure why a $\pi$ system is enough, shouldn't there be closure under (relative) compliments as well? After all, {open sets not containing 0} are a $\pi$ -system, are they not? It seems like you're appealing to the Caratheodory theorem, but that needs the compliment bit as well, or I'm misunderstanding where you're coming from on that.

Oops, I don't mean $\sigma$ -finiteness on $(X,\mathcal{B})$ , but $\sigma$ -finite using sets in the pi-system $\mathcal{Q}$ , i.e. there exists $X_i\in\mathcal{Q}$ such that $\bigcup X_i=X$ (that disqualifies the {open sets not containing 0} example because you can never cover 0).

$\sigma$ -finiteness allows us to add up and subtract, so we can get relative complements:
$\mu(X_i-X_j)=\mu(X_i)-\mu(X_i\cap X_j)$ etc (so you can enlarge $\mathcal{Q}$ by assuming X_i

are nested) and
$\mu(A-B)=\lim_i(\mu(A\cap X_i)-\mu(B\cap X_i))$ (for $B\subset A$ ).

I'm just a little confused though, what does $\sigma$ -finiteness have to do with whether or not the $\pi$ -system covers the space? I get what you meant to say now, this is just a question on terminology.

Ah, OK, then my hunch was right. Thank you very much for confirming that.
It has nothing to do there, except I was thinking about the general statement of uniqueness of measure extended from pi-system to its sigma-algebra:

Theorem: Let $\lambda ,\mu$ be measures on a measurable space $(X,\mathcal{A})$ . Suppose that

is a $\pi$ -system on

generating $\mathcal{A}$ such that $\lambda=\mu$ on

and that there exists a sequence $S_n\in A$ with $\bigcup_{n=1}^\infty S_n=X$ and $\lambda(S_n)<\infty$ . Then $\lambda=\mu$ .
Proof: Apply Dynkin's $\pi\text{-}\lambda$ theorem to $\{S\in\mathcal{A}\mid\lambda(S)=\mu(S)\}$ . Details omitted.

The non-uniqueness when you don't have S_n

is demonstrated by $X=\mathbb{Q}$ , $A=\{[a,b)\cap\mathbb{Q}\mid a,b\in\mathbb{Q}\}$ , when any multiple of counting measure will give the same result (infinity on nonempty) on

.

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Measure preserving

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