Why the complementarity property of $sigma$-algebras?Field of sets and Sigma algebra of setsSigma-algebra...
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Why the complementarity property of $sigma$-algebras?
Field of sets and Sigma algebra of setsSigma-algebra requirement 3, closed under countable unions.Wondering if something is an algebra. If it is, question about closure under complements.$sigma$ additivity$sigma$-algebra of sets that are both $G_delta$ and $F_sigma$How does the $sigma$-algebra axioms ensure meaningful measures $mu$?Measurable sets in sigma algebra generated by $A$sigma algebra v.s. topologyInner and outer measureFor finite $sigma$-algebras, to check closure of countable unions, does it suffice to show that the union of two events is also an event?
$begingroup$
I understand why we limit measures to be defined on domains closed-under-countable union. If we’re going to require countable additivity, we might as well have countable unions belong to the domain.
But why do we limit measures to be defined on domains closed-under-complement?
Possible answers:
Answer: Because in probability, we definitely want closure-under-complement.
Response: But that’s an argument for requiring closure-under-complement for probability measures, not all measures.
Answer: Because given with the other $sigma$-algebra axioms, closure-under-complement implies closure-under-intersection. We definitely want closure-under-intersection, because otherwise every time we referred to an intersection, we’d have to add a disclaimer about the intersection being measurable.
Response: But then why not just require the domains of measures to be closed-under-union and closed-under-intersection? It would make for a natural symmetry! What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
probability-theory measure-theory
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
add a comment |
$begingroup$
I understand why we limit measures to be defined on domains closed-under-countable union. If we’re going to require countable additivity, we might as well have countable unions belong to the domain.
But why do we limit measures to be defined on domains closed-under-complement?
Possible answers:
Answer: Because in probability, we definitely want closure-under-complement.
Response: But that’s an argument for requiring closure-under-complement for probability measures, not all measures.
Answer: Because given with the other $sigma$-algebra axioms, closure-under-complement implies closure-under-intersection. We definitely want closure-under-intersection, because otherwise every time we referred to an intersection, we’d have to add a disclaimer about the intersection being measurable.
Response: But then why not just require the domains of measures to be closed-under-union and closed-under-intersection? It would make for a natural symmetry! What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
probability-theory measure-theory
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15
add a comment |
$begingroup$
I understand why we limit measures to be defined on domains closed-under-countable union. If we’re going to require countable additivity, we might as well have countable unions belong to the domain.
But why do we limit measures to be defined on domains closed-under-complement?
Possible answers:
Answer: Because in probability, we definitely want closure-under-complement.
Response: But that’s an argument for requiring closure-under-complement for probability measures, not all measures.
Answer: Because given with the other $sigma$-algebra axioms, closure-under-complement implies closure-under-intersection. We definitely want closure-under-intersection, because otherwise every time we referred to an intersection, we’d have to add a disclaimer about the intersection being measurable.
Response: But then why not just require the domains of measures to be closed-under-union and closed-under-intersection? It would make for a natural symmetry! What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
probability-theory measure-theory
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
I understand why we limit measures to be defined on domains closed-under-countable union. If we’re going to require countable additivity, we might as well have countable unions belong to the domain.
But why do we limit measures to be defined on domains closed-under-complement?
Possible answers:
Answer: Because in probability, we definitely want closure-under-complement.
Response: But that’s an argument for requiring closure-under-complement for probability measures, not all measures.
Answer: Because given with the other $sigma$-algebra axioms, closure-under-complement implies closure-under-intersection. We definitely want closure-under-intersection, because otherwise every time we referred to an intersection, we’d have to add a disclaimer about the intersection being measurable.
Response: But then why not just require the domains of measures to be closed-under-union and closed-under-intersection? It would make for a natural symmetry! What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
probability-theory measure-theory
probability-theory measure-theory
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
edited Mar 12 at 19:19
Yatharth Agarwal
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 4:12
Yatharth AgarwalYatharth Agarwal
542418
542418
$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15
add a comment |
$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15
$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15
$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Without complementarity you wouldn’t be able to talk about intersections without prefacing every statement involving an intersection that the intersection is measurable. As far as probability theory is concerned, you wouldn’t be able to talk about independence. More fundamentally, how else would you integrate a function like $a1_{X} + b1_{Asetminus X}$?
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
$endgroup$
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
|
show 1 more comment
$begingroup$
What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
Consider measure space $(F, mathcal F, mu)$. For any two subsets of $F$, say $A$ and $B$, it is true that you can get measurability of $A cup B$ using closure-under-intersection. However, you don’t get measurability of $A setminus B$, $B setminus A$, or even $A oplus B$ (symmetric difference)—quite unsatisfactory!
When we know $mu(A cap B)$ and $mu(A cup B)$, and we have countable additivity and $(A oplus B)$ and $(A cap B)$ partition $(A cup B)$, it’s only natural to want $mu(A oplus B)$ to be defined and to be equal to $mu(A cup B) - mu(A cap B)$.
Closure-under-complement gives us that. Closure-under-intersection doesn’t.
If you wanted, the $sigma$-algebra properties could equivalently be including $varnothing$, closure-under-complement, and closure-under-intersection, since the latter two together give us closure-under-union. In fact, you could even have closure-under-complement, closure-under-union, and closure-under-intersection. Closure-under-complement really is the star of the show here, with any 2 of the remaining three being enough.
This should not be a surprising result. Both closure-under-union and closure-under-intersection are two-set axioms, whereas closure-under-complement is a one-set axiom. To describe all parts of the Venn diagram, need the one-set axiom and either two-set axiom. Given the one-set axiom and either two-set axiom, the other axiom becomes redundant. The two-set axioms together don’t give you the one-set axiom.
There might be other reasons as well. For example, closure-under-complement (along with the empty set being included) implies that $mu(A)$ is defined. That’s an important implication for things like Gaussian integrals to be defined, though this line of reasoning is less convincing to me.
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Without complementarity you wouldn’t be able to talk about intersections without prefacing every statement involving an intersection that the intersection is measurable. As far as probability theory is concerned, you wouldn’t be able to talk about independence. More fundamentally, how else would you integrate a function like $a1_{X} + b1_{Asetminus X}$?
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
$endgroup$
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
|
show 1 more comment
$begingroup$
Without complementarity you wouldn’t be able to talk about intersections without prefacing every statement involving an intersection that the intersection is measurable. As far as probability theory is concerned, you wouldn’t be able to talk about independence. More fundamentally, how else would you integrate a function like $a1_{X} + b1_{Asetminus X}$?
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
$endgroup$
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
|
show 1 more comment
$begingroup$
Without complementarity you wouldn’t be able to talk about intersections without prefacing every statement involving an intersection that the intersection is measurable. As far as probability theory is concerned, you wouldn’t be able to talk about independence. More fundamentally, how else would you integrate a function like $a1_{X} + b1_{Asetminus X}$?
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
$endgroup$
Without complementarity you wouldn’t be able to talk about intersections without prefacing every statement involving an intersection that the intersection is measurable. As far as probability theory is concerned, you wouldn’t be able to talk about independence. More fundamentally, how else would you integrate a function like $a1_{X} + b1_{Asetminus X}$?
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
answered Mar 12 at 8:28
CalculonCalculon
3,1101723
3,1101723
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
|
show 1 more comment
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
I understand how closure under complement implies closure under countable intersection. But.. why not just make the property closure under countable intersection in the first place? It would very naturally pair with the closure under countable union.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:14
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
At the end of the day, sure, mathematical objects are constructed arbitrarily, and we could have separate names for what I describe with closure under intersection and for closure under complement. However, I do think there might exist some insight I'm lacking into why X being in the domain of mu was so important as for measures to be universally chosen this way.
$endgroup$
– Yatharth Agarwal
Mar 12 at 10:16
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@YatharthAgarwal Complement + union implies intersection.
$endgroup$
– Calculon
Mar 12 at 12:15
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
@Calulon Right, that's what I meant when I said I understand closure under complement implies complement under union. But why not have the two axioms be complement under union and complement under intersections?
$endgroup$
– Yatharth Agarwal
Mar 12 at 18:45
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
$begingroup$
I ended up answering my own question; posted a self-answer below. Accepting/+1’ing—thank you for your answer!
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:57
|
show 1 more comment
$begingroup$
What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
Consider measure space $(F, mathcal F, mu)$. For any two subsets of $F$, say $A$ and $B$, it is true that you can get measurability of $A cup B$ using closure-under-intersection. However, you don’t get measurability of $A setminus B$, $B setminus A$, or even $A oplus B$ (symmetric difference)—quite unsatisfactory!
When we know $mu(A cap B)$ and $mu(A cup B)$, and we have countable additivity and $(A oplus B)$ and $(A cap B)$ partition $(A cup B)$, it’s only natural to want $mu(A oplus B)$ to be defined and to be equal to $mu(A cup B) - mu(A cap B)$.
Closure-under-complement gives us that. Closure-under-intersection doesn’t.
If you wanted, the $sigma$-algebra properties could equivalently be including $varnothing$, closure-under-complement, and closure-under-intersection, since the latter two together give us closure-under-union. In fact, you could even have closure-under-complement, closure-under-union, and closure-under-intersection. Closure-under-complement really is the star of the show here, with any 2 of the remaining three being enough.
This should not be a surprising result. Both closure-under-union and closure-under-intersection are two-set axioms, whereas closure-under-complement is a one-set axiom. To describe all parts of the Venn diagram, need the one-set axiom and either two-set axiom. Given the one-set axiom and either two-set axiom, the other axiom becomes redundant. The two-set axioms together don’t give you the one-set axiom.
There might be other reasons as well. For example, closure-under-complement (along with the empty set being included) implies that $mu(A)$ is defined. That’s an important implication for things like Gaussian integrals to be defined, though this line of reasoning is less convincing to me.
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
add a comment |
$begingroup$
What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
Consider measure space $(F, mathcal F, mu)$. For any two subsets of $F$, say $A$ and $B$, it is true that you can get measurability of $A cup B$ using closure-under-intersection. However, you don’t get measurability of $A setminus B$, $B setminus A$, or even $A oplus B$ (symmetric difference)—quite unsatisfactory!
When we know $mu(A cap B)$ and $mu(A cup B)$, and we have countable additivity and $(A oplus B)$ and $(A cap B)$ partition $(A cup B)$, it’s only natural to want $mu(A oplus B)$ to be defined and to be equal to $mu(A cup B) - mu(A cap B)$.
Closure-under-complement gives us that. Closure-under-intersection doesn’t.
If you wanted, the $sigma$-algebra properties could equivalently be including $varnothing$, closure-under-complement, and closure-under-intersection, since the latter two together give us closure-under-union. In fact, you could even have closure-under-complement, closure-under-union, and closure-under-intersection. Closure-under-complement really is the star of the show here, with any 2 of the remaining three being enough.
This should not be a surprising result. Both closure-under-union and closure-under-intersection are two-set axioms, whereas closure-under-complement is a one-set axiom. To describe all parts of the Venn diagram, need the one-set axiom and either two-set axiom. Given the one-set axiom and either two-set axiom, the other axiom becomes redundant. The two-set axioms together don’t give you the one-set axiom.
There might be other reasons as well. For example, closure-under-complement (along with the empty set being included) implies that $mu(A)$ is defined. That’s an important implication for things like Gaussian integrals to be defined, though this line of reasoning is less convincing to me.
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
add a comment |
$begingroup$
What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
Consider measure space $(F, mathcal F, mu)$. For any two subsets of $F$, say $A$ and $B$, it is true that you can get measurability of $A cup B$ using closure-under-intersection. However, you don’t get measurability of $A setminus B$, $B setminus A$, or even $A oplus B$ (symmetric difference)—quite unsatisfactory!
When we know $mu(A cap B)$ and $mu(A cup B)$, and we have countable additivity and $(A oplus B)$ and $(A cap B)$ partition $(A cup B)$, it’s only natural to want $mu(A oplus B)$ to be defined and to be equal to $mu(A cup B) - mu(A cap B)$.
Closure-under-complement gives us that. Closure-under-intersection doesn’t.
If you wanted, the $sigma$-algebra properties could equivalently be including $varnothing$, closure-under-complement, and closure-under-intersection, since the latter two together give us closure-under-union. In fact, you could even have closure-under-complement, closure-under-union, and closure-under-intersection. Closure-under-complement really is the star of the show here, with any 2 of the remaining three being enough.
This should not be a surprising result. Both closure-under-union and closure-under-intersection are two-set axioms, whereas closure-under-complement is a one-set axiom. To describe all parts of the Venn diagram, need the one-set axiom and either two-set axiom. Given the one-set axiom and either two-set axiom, the other axiom becomes redundant. The two-set axioms together don’t give you the one-set axiom.
There might be other reasons as well. For example, closure-under-complement (along with the empty set being included) implies that $mu(A)$ is defined. That’s an important implication for things like Gaussian integrals to be defined, though this line of reasoning is less convincing to me.
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
What else besides closure-under-intersection does closure-under-complement imply that’s so important as to be required for every single measure?
Consider measure space $(F, mathcal F, mu)$. For any two subsets of $F$, say $A$ and $B$, it is true that you can get measurability of $A cup B$ using closure-under-intersection. However, you don’t get measurability of $A setminus B$, $B setminus A$, or even $A oplus B$ (symmetric difference)—quite unsatisfactory!
When we know $mu(A cap B)$ and $mu(A cup B)$, and we have countable additivity and $(A oplus B)$ and $(A cap B)$ partition $(A cup B)$, it’s only natural to want $mu(A oplus B)$ to be defined and to be equal to $mu(A cup B) - mu(A cap B)$.
Closure-under-complement gives us that. Closure-under-intersection doesn’t.
If you wanted, the $sigma$-algebra properties could equivalently be including $varnothing$, closure-under-complement, and closure-under-intersection, since the latter two together give us closure-under-union. In fact, you could even have closure-under-complement, closure-under-union, and closure-under-intersection. Closure-under-complement really is the star of the show here, with any 2 of the remaining three being enough.
This should not be a surprising result. Both closure-under-union and closure-under-intersection are two-set axioms, whereas closure-under-complement is a one-set axiom. To describe all parts of the Venn diagram, need the one-set axiom and either two-set axiom. Given the one-set axiom and either two-set axiom, the other axiom becomes redundant. The two-set axioms together don’t give you the one-set axiom.
There might be other reasons as well. For example, closure-under-complement (along with the empty set being included) implies that $mu(A)$ is defined. That’s an important implication for things like Gaussian integrals to be defined, though this line of reasoning is less convincing to me.
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
edited Mar 12 at 20:42
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
answered Mar 12 at 19:56
Yatharth AgarwalYatharth Agarwal
542418
542418
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
add a comment |
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
When you say "it’s only natural to want ...", that is pretty much the only answer one can give to this question. I was pointing out the same sentiment with the example that I gave.
$endgroup$
– Calculon
Mar 12 at 20:04
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
$begingroup$
@Calculon Definitely. What I was missing is understanding the gravity of a particular implication to see why it’s natural to want it.
$endgroup$
– Yatharth Agarwal
Mar 12 at 20:08
add a comment |
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$begingroup$
Note: Before my last edit, I assumed in my question that (given closure-under-union and $mu(varnothing)=0$ being defined), closure-under-complement could be reduced to requiring (a) closure-under-intersection and (b) $mu(A)$ being defined $implies$. However, that was wrong. Closure-under-complement has other consequences, as I detail in my self-answer below.
$endgroup$
– Yatharth Agarwal
Mar 12 at 19:15