Mixing
Terrence Tao's definition of Lebesgue measurability as an extension of Jordan measurability
$begingroup$
I'm reading Terrence Tao's An Introduction to Measure Theory. On pp. 20, he writes,
In analogy with the Jordan theory, we would also like to define
a concept of “Lebesgue inner measure” to complement that of outer
measure. Here, there is an asymmetry (which ultimately arises from
the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner
measure by replacing finite unions of boxes with countable ones.
I don't quite get this argument. The definition of the Jordan Inner Measure is:
For a bounded set, $A$, in $mathbb{R}^d$,
begin{align}
m_{* (J)}(A) = sup_{E subset A} m(A),
end{align}
where $E$ is an elementary set in $mathbb{R}^d$ definite to be the finite union of boxes, which are just $d-$dimensional Cartesian products of intervals in $mathbb{R}$.
If we replace finite unions by countable unions, we can define,
begin{align}
m_{* (L)}(A) = sup_{E subset A} m(A),
end{align}
where $m_{*(L)}(A)$ is the Lebesgue inner measure.
Clearly, we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A).
end{align}
This is analogous to the result,
begin{align}
m^{*(L)}(A) leq m^{*(J)}(A).
end{align}
Since it's not too hard to show that,
begin{align}
m_{*(L)}(A) leq m^{*(L)}(A),
end{align}
we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A) leq m^{*(L)}(A) leq m^{*(L)}(A).
end{align}
Hence the Lebesgue measure can be though of as a refinement of the Jordan measure.
In light of this discussion, why does Tao dismiss the idea of defining the Lebesgue measurable sets as those for which the inner and outer measures coincide?
Here are the link to the notes:
https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
measure-theory
asked Jan 3 at 15:50
user82261user82261
20217
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|
show 3 more comments
$begingroup$
I'm reading Terrence Tao's An Introduction to Measure Theory. On pp. 20, he writes,
In analogy with the Jordan theory, we would also like to define
a concept of “Lebesgue inner measure” to complement that of outer
measure. Here, there is an asymmetry (which ultimately arises from
the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner
measure by replacing finite unions of boxes with countable ones.
I don't quite get this argument. The definition of the Jordan Inner Measure is:
For a bounded set, $A$, in $mathbb{R}^d$,
begin{align}
m_{* (J)}(A) = sup_{E subset A} m(A),
end{align}
where $E$ is an elementary set in $mathbb{R}^d$ definite to be the finite union of boxes, which are just $d-$dimensional Cartesian products of intervals in $mathbb{R}$.
If we replace finite unions by countable unions, we can define,
begin{align}
m_{* (L)}(A) = sup_{E subset A} m(A),
end{align}
where $m_{*(L)}(A)$ is the Lebesgue inner measure.
Clearly, we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A).
end{align}
This is analogous to the result,
begin{align}
m^{*(L)}(A) leq m^{*(J)}(A).
end{align}
Since it's not too hard to show that,
begin{align}
m_{*(L)}(A) leq m^{*(L)}(A),
end{align}
we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A) leq m^{*(L)}(A) leq m^{*(L)}(A).
end{align}
Hence the Lebesgue measure can be though of as a refinement of the Jordan measure.
In light of this discussion, why does Tao dismiss the idea of defining the Lebesgue measurable sets as those for which the inner and outer measures coincide?
Here are the link to the notes:
https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
measure-theory
asked Jan 3 at 15:50
user82261user82261
20217
$endgroup$
$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15
|
show 3 more comments
$begingroup$
I'm reading Terrence Tao's An Introduction to Measure Theory. On pp. 20, he writes,
In analogy with the Jordan theory, we would also like to define
a concept of “Lebesgue inner measure” to complement that of outer
measure. Here, there is an asymmetry (which ultimately arises from
the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner
measure by replacing finite unions of boxes with countable ones.
I don't quite get this argument. The definition of the Jordan Inner Measure is:
For a bounded set, $A$, in $mathbb{R}^d$,
begin{align}
m_{* (J)}(A) = sup_{E subset A} m(A),
end{align}
where $E$ is an elementary set in $mathbb{R}^d$ definite to be the finite union of boxes, which are just $d-$dimensional Cartesian products of intervals in $mathbb{R}$.
If we replace finite unions by countable unions, we can define,
begin{align}
m_{* (L)}(A) = sup_{E subset A} m(A),
end{align}
where $m_{*(L)}(A)$ is the Lebesgue inner measure.
Clearly, we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A).
end{align}
This is analogous to the result,
begin{align}
m^{*(L)}(A) leq m^{*(J)}(A).
end{align}
Since it's not too hard to show that,
begin{align}
m_{*(L)}(A) leq m^{*(L)}(A),
end{align}
we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A) leq m^{*(L)}(A) leq m^{*(L)}(A).
end{align}
Hence the Lebesgue measure can be though of as a refinement of the Jordan measure.
In light of this discussion, why does Tao dismiss the idea of defining the Lebesgue measurable sets as those for which the inner and outer measures coincide?
Here are the link to the notes:
https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
measure-theory
asked Jan 3 at 15:50
user82261user82261
20217
$endgroup$
I'm reading Terrence Tao's An Introduction to Measure Theory. On pp. 20, he writes,
In analogy with the Jordan theory, we would also like to define
a concept of “Lebesgue inner measure” to complement that of outer
measure. Here, there is an asymmetry (which ultimately arises from
the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner
measure by replacing finite unions of boxes with countable ones.
I don't quite get this argument. The definition of the Jordan Inner Measure is:
For a bounded set, $A$, in $mathbb{R}^d$,
begin{align}
m_{* (J)}(A) = sup_{E subset A} m(A),
end{align}
where $E$ is an elementary set in $mathbb{R}^d$ definite to be the finite union of boxes, which are just $d-$dimensional Cartesian products of intervals in $mathbb{R}$.
If we replace finite unions by countable unions, we can define,
begin{align}
m_{* (L)}(A) = sup_{E subset A} m(A),
end{align}
where $m_{*(L)}(A)$ is the Lebesgue inner measure.
Clearly, we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A).
end{align}
This is analogous to the result,
begin{align}
m^{*(L)}(A) leq m^{*(J)}(A).
end{align}
Since it's not too hard to show that,
begin{align}
m_{*(L)}(A) leq m^{*(L)}(A),
end{align}
we have,
begin{align}
m_{*(J)}(A) leq m_{*(L)}(A) leq m^{*(L)}(A) leq m^{*(L)}(A).
end{align}
Hence the Lebesgue measure can be though of as a refinement of the Jordan measure.
In light of this discussion, why does Tao dismiss the idea of defining the Lebesgue measurable sets as those for which the inner and outer measures coincide?
Here are the link to the notes:
https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
measure-theory
measure-theory
asked Jan 3 at 15:50
user82261user82261
20217
asked Jan 3 at 15:50
user82261user82261
20217
edited Jan 3 at 16:01
user82261
asked Jan 3 at 15:50
user82261user82261
20217
asked Jan 3 at 15:50
user82261user82261
20217
asked Jan 3 at 15:50
user82261user82261
20217
20217
$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15
|
show 3 more comments
$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15
$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15
|
show 3 more comments
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$begingroup$
Extending the definition of inner Jordan content to countable collections of elementary boxes doesn't change the value of the inner content on bounded open sets and because an open elementary set contained in an arbitrary set $E$ is necessarily contained in $E^{circ}$, the same is true for $any$ bounded set $E$.
$endgroup$
– Matematleta
Jan 3 at 16:49
$begingroup$
@Matematleta Can you elaborate on how and why is the first statement true? Perhaps in the form of an answer. I didn't play around with any specific sets to see how the definition turns out, so I never really thought of it.
$endgroup$
– user82261
Jan 3 at 17:01
$begingroup$
@Matematleta I get that the inner Jordan content of any set, $E$, is the same as its interior. But I don't get how extending the definition to inner Lebesgue measure would not change in general the inner measure of open sets, and, consequently as you argue, of all bounded sets.
$endgroup$
– user82261
Jan 3 at 17:04
$begingroup$
@Matematleta: P.S, the fact the innter Jordan content of any set is the same as that of its interior, and the outer Jordan content of any set is the same as that of its closure is something that wasn't, by all means, obvious to me. So I suppose I'm failing to make some connection.
$endgroup$
– user82261
Jan 3 at 17:11
$begingroup$
Every open set in $mathbb R^n$ is a countable disjoint union of elementary boxes, so the inner content can be approximated to any desired degree by a suitable finite collection of boxes. Here is a link that may help with how the inner and outer Jordan content are related.
$endgroup$
– Matematleta
Jan 3 at 17:15