A $sigma$-algebra is a monotone class.
$begingroup$
I am having trouble understanding the first line of a proof.
"Suppose $A_0$ is an algebra, $A$ is the smallest $sigma$-algebra containing $A_0$, and $M$ is the smallest monotone class containing $A_0$, The $M=A$.
Proof.
A $sigma$-algebra is clearly a monotone class, so $M subset A$."
I would think that we would have the opposite inclusion. A truck is a car, so {Trucks} $subset$ {Cars}?
real-analysis
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding the first line of a proof.
"Suppose $A_0$ is an algebra, $A$ is the smallest $sigma$-algebra containing $A_0$, and $M$ is the smallest monotone class containing $A_0$, The $M=A$.
Proof.
A $sigma$-algebra is clearly a monotone class, so $M subset A$."
I would think that we would have the opposite inclusion. A truck is a car, so {Trucks} $subset$ {Cars}?
real-analysis
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
$endgroup$
3
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31
add a comment |
$begingroup$
I am having trouble understanding the first line of a proof.
"Suppose $A_0$ is an algebra, $A$ is the smallest $sigma$-algebra containing $A_0$, and $M$ is the smallest monotone class containing $A_0$, The $M=A$.
Proof.
A $sigma$-algebra is clearly a monotone class, so $M subset A$."
I would think that we would have the opposite inclusion. A truck is a car, so {Trucks} $subset$ {Cars}?
real-analysis
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
$endgroup$
I am having trouble understanding the first line of a proof.
"Suppose $A_0$ is an algebra, $A$ is the smallest $sigma$-algebra containing $A_0$, and $M$ is the smallest monotone class containing $A_0$, The $M=A$.
Proof.
A $sigma$-algebra is clearly a monotone class, so $M subset A$."
I would think that we would have the opposite inclusion. A truck is a car, so {Trucks} $subset$ {Cars}?
real-analysis
real-analysis
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
asked Jan 8 at 22:06
JungleshrimpJungleshrimp
366112
366112
3
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31
add a comment |
3
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31
3
3
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31
add a comment |
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3
$begingroup$
...Hence the smallest truck A is larger than the smallest car M. QED.
$endgroup$
– Did
Jan 8 at 22:16
$begingroup$
I don't get it...
$endgroup$
– Jungleshrimp
Jan 8 at 22:19
$begingroup$
By your own remark-analogy, {Trucks} ⊂ {Cars} and M = inf {Cars} and A = inf {Trucks}, hence you are done.
$endgroup$
– Did
Jan 8 at 22:22
$begingroup$
$A$ is sigma algebra, so it's a monotone class. $M$ is the $smallest$ monotone class that contains $A_0$. Hence...
$endgroup$
– Matematleta
Jan 8 at 22:24
$begingroup$
Thank you both.
$endgroup$
– Jungleshrimp
Jan 8 at 22:31