\(
\newcommand{\cat}[1] {\mathrm{#1}}
\newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})}
\newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}}
\)
Math and science::Analysis::Tao, measure
A motivation of measure [stub]
[...] is a motivational progression though different ways of describing and
measuring the volume of things in \( R^d \) space.
The standard integral
A lot of people feel an intuitive understanding when they see something like
the expression \( A = \int_{a}^{b} f(x) dx \) and a figure like:
TODO: insert figure.
When I look at the expression I interpret \( A \) as the "area" under the
graph, where my concept of "area" is an intuitive one, rather than an a concept
arising from some mathematical definition.
Integration in the form of a challenge-response exchange
The following exchange starts to approach more closely what \( A \) really
represents mathematically.
Imagine wise looking man approaches you as you walk down the street. His
name is Darboux and wants to show you his trick. He has a special way of filling
and covering things with rectangles. Specifically, he shows you a figure [TODO,
show x-y function]. And he says, for this figure he has a special number,
7.214.
He challenges you: no matter what positive number \( \varepsilon > 0 \)
you choose, he can do two things:
- He can find a set of non-overlapping rectangles (except possibly for
their edges) that fit inside \( A \), and these rectangles will have a combined
area (sum of \( \text{height} \times \text{width} \)) that is greater than
\( A - \varepsilon \).
- I can find a different set of non-overlapping rectangles (except
possibly for their edges) that contain \( A \), and these rectangles will have
a combined area (sum of \( \text{height} \times \{width} \)) that is less than
\( A - \varepsilon \).
Furthermore, he says, he doesn't need infinite rectangles, he will do it
with a finite set of rectangles: for each \( \varepsilon > 0 \) you give, he
can tell you the number \( N \) of rectangles he needs in order to fill or cover
the function.
After he convinces you of his ability, he then challenges you! Try find
another number other than 7.214 and see if you can fill and cover the function
with rectangles like above. Better to not take up this challenge, for it is
impossible.
This is a challenge response interpretation of the limit formulation of
the classic integral.
Shapes in \( R^d \)
Darboux could only fill and cover the space under a function
\( \mathbb{R} \to \mathbb{R} \). A few months later, as you are walking down the
same street as before, a different wise looking man approaches you. His name
is Jordan, and he too wants to show you a trick. He said that Darboux's trick
was weak in comparison to his trick! Not limited to filling and covering the
space under a function, Jordan says: "pick any shape in 3D or any
\( \mathbb{R}^d \) Euclidean space and I will find a special number \( A \in
\mathbb{R} \) and two sets of rectangles, one that covers and one that is
contained within the shape. Like before, there won't be an infinite number of
rectangles, they won't overlap (except possibly the edges) and the volume of the
rectangles in each set will sum to the special number \( A \).
The breakdown of Jordan measure 1
Jordan measure fails to give measure to countable unions or intersections
of sets that are already known to be measurable. The bullets are an example.
The breakdown of Jordan measure 2
Imagine using the Jordan measure to determine probabilities. When
requested for a probability, he convert the problem into a geometric
representation in \( \mathbb{R}^d \) then calculate the "volume"/
"measure" of both the space representing all possible events and the "volume"
of an event in question.
However, here is a case where the Jordan measure will breakdown. Imagine
a random number generator that generates real numbers between 1 and 2. What is
the probability that the number generated is a rational and what is the
probability that the number generated is an irrational? This can be though of as
a question about the relative abundance of rationals and irrationals within
\( [1, 2] \).
Jordan's approach is to calculate the "volume" of the rationals: try to find
two sets of disjoint rectangles, one set of that covers the rationals and one
that is contained within the rationals. But the only boxes that fit within the
rationals have zero volume. For example, \( (\frac{4}{3} \) and
\( \frac{4}{3}) \), so the set of boxes that are contained within the rationals
cannot have a combined volume greater than zero. This would suggest that
Jordan's measure is 0. However, when we investigate the outer cover, we
encounter a problem. If we restrict ourselves to finite sets of boxes, we can't
cover the rationals with disjoint boxes with a combined volume less that 1.
[TODO: tidy up terminology of boxes and intervals and rectangles]
So we have two different numbers for our volume measurement: 0 or 1. If we
relate these back to the question of probability, this means that the random
number generator will either never or always produce rationals.
Lebesgue measure
The challenge-response interpretation of Jordan measure for a shape involved
choosing an \( \varepsilon > 0 \) and having Jordan find a two finite sets of
boxes--one set that covers the shape and one that is contained within the shape
in such a way that both box covers have the same volume when calculated as the
sum of the volumes of each box.
The Lebesgue measure relaxes the requirement for the box cover to be
finite. Given an \( \varepsilon > 0 \), Lebesgue can find a countable set of
boxes that covers a shape.
Darboux vs Riemann integral
The Darboux integral and the Riemann integral are two equivalent ways of
formulating the classical integral. They are equivalent in the sense that they
always have the same value (when they work), and they break down in the same
circumstances. It's as if they implement the same programming interface and
always produce the same output for a given input. They differ in how they
define the integral.
The Darboux integral has two parts: an upper and lower
integral. The upper is the infimum of volume for almost-disjoint rectangle
covers of the area under a function. The lower is the supremum of volume
for almost-disjoint sets of rectangles contained within the space under a
function. When the supremum and infimum are equal,
or the Riemann integral)
Transition to measure any shape (Jordan measure)
Jordan measurability
The most interesting thing: what is not considered "measurable" an why?
Transition to uncountable boxes
Measurability again
\(
\newcommand{\cat}[1] {\mathrm{#1}}
\newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})}
\newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}}
\)
Math and science::Analysis::Tao, measure
A motivation of measure [stub]
This is a motivational progression though different ways of describing and
measuring the volume of things in \( R^d \) space.
The standard integral
A lot of people feel an intuitive understanding when they see something like
the expression \( A = \int_{a}^{b} f(x) dx \) and a figure like:
TODO: insert figure.
When I look at the expression I interpret \( A \) as the "area" under the
graph, where my concept of "area" is an intuitive one, rather than an a concept
arising from some mathematical definition.
Integration in the form of a challenge-response exchange
The following exchange starts to approach more closely what \( A \) really
represents mathematically.
Imagine wise looking man approaches you as you walk down the street. His
name is Darboux and wants to show you his trick. He has a special way of filling
and covering things with rectangles. Specifically, he shows you a figure [TODO,
show x-y function]. And he says, for this figure he has a special number,
7.214.
He challenges you: no matter what positive number \( \varepsilon > 0 \)
you choose, he can do two things:
- He can find a set of non-overlapping rectangles (except possibly for
their edges) that fit inside \( A \), and these rectangles will have a combined
area (sum of \( \text{height} \times \text{width} \)) that is greater than
\( A - \varepsilon \).
- I can find a different set of non-overlapping rectangles (except
possibly for their edges) that contain \( A \), and these rectangles will have
a combined area (sum of \( \text{height} \times \{width} \)) that is less than
\( A - \varepsilon \).
Furthermore, he says, he doesn't need infinite rectangles, he will do it
with a finite set of rectangles: for each \( \varepsilon > 0 \) you give, he
can tell you the number \( N \) of rectangles he needs in order to fill or cover
the function.
After he convinces you of his ability, he then challenges you! Try find
another number other than 7.214 and see if you can fill and cover the function
with rectangles like above. Better to not take up this challenge, for it is
impossible.
This is a challenge response interpretation of the limit formulation of
the classic integral.
Shapes in \( R^d \)
Darboux could only fill and cover the space under a function
\( \mathbb{R} \to \mathbb{R} \). A few months later, as you are walking down the
same street as before, a different wise looking man approaches you. His name
is Jordan, and he too wants to show you a trick. He said that Darboux's trick
was weak in comparison to his trick! Not limited to filling and covering the
space under a function, Jordan says: "pick any shape in 3D or any
\( \mathbb{R}^d \) Euclidean space and I will find a special number \( A \in
\mathbb{R} \) and two sets of rectangles, one that covers and one that is
contained within the shape. Like before, there won't be an infinite number of
rectangles, they won't overlap (except possibly the edges) and the volume of the
rectangles in each set will sum to the special number \( A \).
The breakdown of Jordan measure 1
Jordan measure fails to give measure to countable unions or intersections
of sets that are already known to be measurable. The bullets are an example.
The breakdown of Jordan measure 2
Imagine using the Jordan measure to determine probabilities. When
requested for a probability, he convert the problem into a geometric
representation in \( \mathbb{R}^d \) then calculate the "volume"/
"measure" of both the space representing all possible events and the "volume"
of an event in question.
However, here is a case where the Jordan measure will breakdown. Imagine
a random number generator that generates real numbers between 1 and 2. What is
the probability that the number generated is a rational and what is the
probability that the number generated is an irrational? This can be though of as
a question about the relative abundance of rationals and irrationals within
\( [1, 2] \).
Jordan's approach is to calculate the "volume" of the rationals: try to find
two sets of disjoint rectangles, one set of that covers the rationals and one
that is contained within the rationals. But the only boxes that fit within the
rationals have zero volume. For example, \( (\frac{4}{3} \) and
\( \frac{4}{3}) \), so the set of boxes that are contained within the rationals
cannot have a combined volume greater than zero. This would suggest that
Jordan's measure is 0. However, when we investigate the outer cover, we
encounter a problem. If we restrict ourselves to finite sets of boxes, we can't
cover the rationals with disjoint boxes with a combined volume less that 1.
[TODO: tidy up terminology of boxes and intervals and rectangles]
So we have two different numbers for our volume measurement: 0 or 1. If we
relate these back to the question of probability, this means that the random
number generator will either never or always produce rationals.
Lebesgue measure
The challenge-response interpretation of Jordan measure for a shape involved
choosing an \( \varepsilon > 0 \) and having Jordan find a two finite sets of
boxes--one set that covers the shape and one that is contained within the shape
in such a way that both box covers have the same volume when calculated as the
sum of the volumes of each box.
The Lebesgue measure relaxes the requirement for the box cover to be
finite. Given an \( \varepsilon > 0 \), Lebesgue can find a countable set of
boxes that covers a shape.
Darboux vs Riemann integral
The Darboux integral and the Riemann integral are two equivalent ways of
formulating the classical integral. They are equivalent in the sense that they
always have the same value (when they work), and they break down in the same
circumstances. It's as if they implement the same programming interface and
always produce the same output for a given input. They differ in how they
define the integral.
The Darboux integral has two parts: an upper and lower
integral. The upper is the infimum of volume for almost-disjoint rectangle
covers of the area under a function. The lower is the supremum of volume
for almost-disjoint sets of rectangles contained within the space under a
function. When the supremum and infimum are equal,
or the Riemann integral)
Transition to measure any shape (Jordan measure)
Jordan measurability
The most interesting thing: what is not considered "measurable" an why?
Transition to uncountable boxes
Measurability again
