Mixing
Show that $mu^#$ (Counting Measure) is $sigma$-finite iff $X$ is countable
Clash Royale CLAN TAG#URR8PPP
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For the '$Rightarrow$' is easy because if we assume $mu^#$ sigma finite then
$$X = bigcup_j=1^infty A_j$$
We have that $mu^#(A_j) lt infty hspace3mmforall j$ so this means that $A_j$ is finite for all j and countable union of finite sets is still countable. I'm having troubles showing the '$Leftarrow$' because if X is countable it could be union of countable sets, so if exists a $j_0$ such that $A_j_0$ is countable as well then how is it possible to have $mu^#(A_j_0) lt infty$?
real-analysis measure-theory
asked Aug 3 at 11:50
James Arten
186
add a comment |Â
up vote
1
down vote
favorite
For the '$Rightarrow$' is easy because if we assume $mu^#$ sigma finite then
$$X = bigcup_j=1^infty A_j$$
We have that $mu^#(A_j) lt infty hspace3mmforall j$ so this means that $A_j$ is finite for all j and countable union of finite sets is still countable. I'm having troubles showing the '$Leftarrow$' because if X is countable it could be union of countable sets, so if exists a $j_0$ such that $A_j_0$ is countable as well then how is it possible to have $mu^#(A_j_0) lt infty$?
real-analysis measure-theory
asked Aug 3 at 11:50
James Arten
186
1
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For the '$Rightarrow$' is easy because if we assume $mu^#$ sigma finite then
$$X = bigcup_j=1^infty A_j$$
We have that $mu^#(A_j) lt infty hspace3mmforall j$ so this means that $A_j$ is finite for all j and countable union of finite sets is still countable. I'm having troubles showing the '$Leftarrow$' because if X is countable it could be union of countable sets, so if exists a $j_0$ such that $A_j_0$ is countable as well then how is it possible to have $mu^#(A_j_0) lt infty$?
real-analysis measure-theory
asked Aug 3 at 11:50
James Arten
186
For the '$Rightarrow$' is easy because if we assume $mu^#$ sigma finite then
$$X = bigcup_j=1^infty A_j$$
We have that $mu^#(A_j) lt infty hspace3mmforall j$ so this means that $A_j$ is finite for all j and countable union of finite sets is still countable. I'm having troubles showing the '$Leftarrow$' because if X is countable it could be union of countable sets, so if exists a $j_0$ such that $A_j_0$ is countable as well then how is it possible to have $mu^#(A_j_0) lt infty$?
real-analysis measure-theory
asked Aug 3 at 11:50
James Arten
186
edited Aug 3 at 11:55
asked Aug 3 at 11:50
James Arten
186
asked Aug 3 at 11:50
James Arten
186
asked Aug 3 at 11:50
James Arten
186
186
1
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55
add a comment |Â
1
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55
1
1
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $Bbb N$ is $bigcup_ninBbb Nn$.
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
add a comment |Â
up vote
2
down vote
To be explicit about something implicit in the existing answer: When you say that $A_j_0$ could be countable it seems that you're confused about the definition of "$sigma$-finite".
Say $mu$ is a measure on $X$. Consider the following four conditions:
(i) $X$ is a countable union of sets of finite measure.
(ii) $X=bigcup_j=1^infty A_j$ where $mu(A_j)<infty$.
(iii) There exist $A_1,A_2,dots$ such that $X=bigcup_j=1^infty A_j$ and $mu(A_j)<infty$.
(iv) If $X=bigcup_j=1^infty A_j$ then $mu(A_j)<infty$.
The definition of "$mu$ is $sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:
(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).
The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.
Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.
(In fact, easy exercise (iv) holds if and only if $mu(X)<infty$.)
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $Bbb N$ is $bigcup_ninBbb Nn$.
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
add a comment |Â
up vote
3
down vote
accepted
It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $Bbb N$ is $bigcup_ninBbb Nn$.
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $Bbb N$ is $bigcup_ninBbb Nn$.
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $Bbb N$ is $bigcup_ninBbb Nn$.
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
answered Aug 3 at 11:55
Asaf Karagila
291k31401731
291k31401731
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
add a comment |Â
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
That's right!! So you mean that I could write my set X as union of singletons and then obviously showing that counting measures of each point will be 1?
– James Arten
Aug 3 at 12:00
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
Well, the counting measure is well known to give a finite set its cardinality. How many elements does a singleton have?
– Asaf Karagila
Aug 3 at 12:01
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
At least one and probably not a whole lot more.
– Drew Brady
4 hours ago
add a comment |Â
up vote
2
down vote
To be explicit about something implicit in the existing answer: When you say that $A_j_0$ could be countable it seems that you're confused about the definition of "$sigma$-finite".
Say $mu$ is a measure on $X$. Consider the following four conditions:
(i) $X$ is a countable union of sets of finite measure.
(ii) $X=bigcup_j=1^infty A_j$ where $mu(A_j)<infty$.
(iii) There exist $A_1,A_2,dots$ such that $X=bigcup_j=1^infty A_j$ and $mu(A_j)<infty$.
(iv) If $X=bigcup_j=1^infty A_j$ then $mu(A_j)<infty$.
The definition of "$mu$ is $sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:
(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).
The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.
Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.
(In fact, easy exercise (iv) holds if and only if $mu(X)<infty$.)
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
add a comment |Â
up vote
2
down vote
To be explicit about something implicit in the existing answer: When you say that $A_j_0$ could be countable it seems that you're confused about the definition of "$sigma$-finite".
Say $mu$ is a measure on $X$. Consider the following four conditions:
(i) $X$ is a countable union of sets of finite measure.
(ii) $X=bigcup_j=1^infty A_j$ where $mu(A_j)<infty$.
(iii) There exist $A_1,A_2,dots$ such that $X=bigcup_j=1^infty A_j$ and $mu(A_j)<infty$.
(iv) If $X=bigcup_j=1^infty A_j$ then $mu(A_j)<infty$.
The definition of "$mu$ is $sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:
(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).
The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.
Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.
(In fact, easy exercise (iv) holds if and only if $mu(X)<infty$.)
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To be explicit about something implicit in the existing answer: When you say that $A_j_0$ could be countable it seems that you're confused about the definition of "$sigma$-finite".
Say $mu$ is a measure on $X$. Consider the following four conditions:
(i) $X$ is a countable union of sets of finite measure.
(ii) $X=bigcup_j=1^infty A_j$ where $mu(A_j)<infty$.
(iii) There exist $A_1,A_2,dots$ such that $X=bigcup_j=1^infty A_j$ and $mu(A_j)<infty$.
(iv) If $X=bigcup_j=1^infty A_j$ then $mu(A_j)<infty$.
The definition of "$mu$ is $sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:
(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).
The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.
Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.
(In fact, easy exercise (iv) holds if and only if $mu(X)<infty$.)
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
To be explicit about something implicit in the existing answer: When you say that $A_j_0$ could be countable it seems that you're confused about the definition of "$sigma$-finite".
Say $mu$ is a measure on $X$. Consider the following four conditions:
(i) $X$ is a countable union of sets of finite measure.
(ii) $X=bigcup_j=1^infty A_j$ where $mu(A_j)<infty$.
(iii) There exist $A_1,A_2,dots$ such that $X=bigcup_j=1^infty A_j$ and $mu(A_j)<infty$.
(iv) If $X=bigcup_j=1^infty A_j$ then $mu(A_j)<infty$.
The definition of "$mu$ is $sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:
(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).
The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.
Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.
(In fact, easy exercise (iv) holds if and only if $mu(X)<infty$.)
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
answered Aug 3 at 15:50
David C. Ullrich
53.8k33481
53.8k33481
add a comment |Â
add a comment |Â
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1
What's $mu^#$?
– Asaf Karagila
Aug 3 at 11:54
Counting measure! Sorry! ^_^
– James Arten
Aug 3 at 11:55