# Lebesgue measure and scale freedom

1. Mar 28, 2008

### my_wan

My skill with proofs and number theory is very limited and my use tends toward the stereotyped physicist. Dealing with infinities and identifying exactly what assumptions are in play has become somewhat an issue for me though. My questions are contained in the following visual tool.
Please slap me in line if I mis-use any terms or logic.

Define two line segments $$l_1$$ and $$l_2$$, y=mx + b, where the slopes are 0.
$$l_1$$ where $$b_1 = 2$$ and $$2<y_1<-2$$
$$l_2$$ where $$b_2 = -1$$ and $$1<y_2<-1$$
Let set $${R_1}$$ be a subset containing all points represented by one and only one real number, cardinality 1. Dito for $$R_2$$.

Theorems:
1) The number of members of the set $$R_1$$ or $$R_2$$ is infinite.
2) Origin bijection - For each line that intercepts $$l_1, l_2,$$ and point (0,0) there exist one and only one member each of $$R_1$$ and $$R_2$$ that this same line intercepts both $$l_1$$ and $$l_2$$, i.e., not surjective.
3) Origin bijection is a real number logical equivalent of the integer bijection used by Cantor in transfinite sets where
$$l_1 \equiv \sum_{n=1}^\infty n$$
and
$$l_2 \equiv \sum_{n=1}^\infty 2n$$
using infinitesimals rather than limits.

Question: What prior art exist similar to this notion of origin bijection?

Implications:
1) Any finite one to one mapping via origin bijection indicates that $$l_2/l_1 = 1$$, yet in the limit must equal 1/2 in this case, and varies generally.
2) Implication 1) indicates the set of infinitesimals not contained in the subset R, which are neither measurable nor countable, must have a total magnitude (summation) equal to the total magnitude of the subset R defined in l.
3) We can maintain that implication 2) is not due to a summation of infinitesimals not contained in the subset R but that indicates that 1 to 1 correspondence is invalid. Vitali sets specifically defines this as a property of reals.
4) We could assume the difference in $$l_1$$ and $$l_2$$ are a result of irrational numbers. This implies that the ratio of irrational numbers to real numbers can take any value, finite and otherwise, i.e., the irrational numbers defined on any line segment can sum to any value we choose simply by choice of scale.
5) Scale freedom (coordinate independence) is a fundamental principle and we could assume that the ratio of reals to irrationals has some actual value that changes due to scale. However, if this was the case, in principle the same bijection effects should apply to the irrationals.
6) If we reject this scale choice as a property of irrationals then we must assume a set of points not defined by any definable number must contain this property of scale freedom, i.e., not Lebesgue measurable. Assuming this is a property of irrationals means that the axiom of choice is the only assumption required to say 1.99... = 2. Defining how and where these non-measurable sets act on scale appears as a moving target and allows choice in how to define them.

Axiom of choice & Lebesgue measure:
Assuming AC not all subsets of a space are Lebesgue measurable. Wrt origin bijection the Lebesgue immeasurability (non-measurable sets) implicit in AC appears as the infinitesimals that are not accounted for in the 1 to 1 origin bijection. This is the property I identified above as scale freedom. This implies that sets which are not Lebesgue measurable may use scale freedom to redefine spaces of one magnitude to spaces of another magnitude. A related case can be made that scale freedom requires this property of non-measurable sets to be mathematically consistent. How would a choice of scale remain mathematically consistant without non-measurable sets with which to define this choice? The non-measurable sets in $$l_1$$ and $$l_2$$ play the same role whether it is a change of scale or simply a redefinition of scale where we simple choose to define length $$l_1 = 1$$.

Conjecture - Lebesgue immeasurable sets and scale freedom or choice is equivalent.

Banach-Tarski theorem & Vitali sets:
Both Banach-Tarski theorem and Vitali sets depend critically on AC and the associated construction of non-measurable sets. Vitali sets define these non-measurable sets as reals yet remain impossible to describe explicitly. If the above conjecture holds then these theorems simply redefines scale rather than construct one volume from another. So the choice in ac that requires non-measurable sets is a choice of scale, not a change of scale, by the above conjecture.

Questions:
Ive seen some references to attempts via nonconstructive proof to exploit these properties of non-measurable sets via AC. What prior work exist that attempts to resolve the paradoxes without rejecting AC? What other theorems etc. might be related to this I should be made aware of? What prior work is similar to the concept of origin bijection I used here? What objections and/or logical errors can be pointed out here?

2. Mar 28, 2008

### ramsey2879

Do you mean that the line segments are parallel to the x axis, i.e. have zero slope and intersect a line with slope m at points y1,x1 and y2,x2 respectively? If not then please clarify since this is the only reasonable interpertation of this statement. A line with zero slope means that m = 0.
If the answer to the first question is yes then I am confused since y1 must equal 2 and y2 must equal -1
How can a point have any numerical meaning, if it is not on a given number line or a predefined grid? Since you have two dimensions then are you talking of complex numbers with real and imaginary parts?

Last edited: Mar 28, 2008
3. Mar 28, 2008

### my_wan

4. Mar 28, 2008

### my_wan

The two lines are parallel. A third line bisects the two parallel lines and point (0,0) for the sole purpose of establishing a 1 to 1 correspondence between a point on $$l_1$$ with a point on $$l_2$$.

A point is defined by the coordinates (x,y), the numerical meaning and the definition of a member of the set R (with cardinality 1).

Essentially it's a method of doing a 1 to 1 correspondence in the same manner Cantor did except on a line segment rather than the whole numbers.

5. Mar 28, 2008

### ramsey2879

The graph that you linked shows that y1 can only equal 2 and y2 can only equal -1. Do you really mean that line l1 and l2 each must have zero slope or do you mean that they have the same slope and the graph is just an example? I still don't understand what you mean by point represented with one and only one real number. Are you talking of the complex number plane?

6. Mar 28, 2008

### ramsey2879

So I take it that the two lines have the same slope but not zero slope since y1 is less than 2 and y2 is greater than -1

7. Mar 28, 2008

### my_wan

The lines have no slope. It's a bijection to show 1 to 1 correspondence between the points that define $$l_1$$ and $$l_2$$. The set R is the set of points that define $$l$$ such as (1,2), (0.5,2), (0.1,2), etc. It's a bijection such that (1,2) corresponds to (0.5,-1) and (0.5,2) corresponds to (0.25,-1), etc. Any point on either line will correspond to one and only one unique point on the other line. An infinite series of points each having a 1 to 1 correspondence to another set of points that in the limit is twice or inversely half the corresponding line segment.

ETA: The slope of $$l_1$$ and $$l_2$$ remain 0 in all cases. It's the bijection line that is incremented on a 1 to 1 basis.

Last edited: Mar 28, 2008
8. Mar 28, 2008

### ramsey2879

Maybe your coordinate systerm is distinct from that in the USA, but in the USA any line passing through the orgin has the equation y = mx + 0 not y = mx + b, now this line passes through the line y = 2 at point x1 and passes through the line y = -1 at the point x2 such that x1 = -2x2. Is that what you are talking about or did I switch the x and y coordinates? By real numbers R1 and R2 you mean the corresponding points on line y = 2 and y = -1 respectively such that the line therebetween passes through the orgin.

With that assumption, yes x and 2x have the same cardinality but that doesn't mean that the points of irrational numbers have varying size. Points on a number line have no size.

9. Mar 28, 2008

### my_wan

In the straight line formula I use (US) y = mx + b where m is the slope and b is the y intercept. It's called the slope-intercept form. It seems you are talking about the point-slope form.

The point wasn't that the irrational numbers have varying sizes, as you said no point has a finite size. Yet a summation of all points at a defined scale sums to a value (size) in principle. We couldn't define a finite line segment as a set of points otherwise. It wasn't even the size of the points defined as irrational I was speaking of but a summation of them across an interval. Nor did I claim it was a property of irrational specifically but that we had choice in how we defined it and related this choice to our choice of scale in defining an interval. Rather than irrationals non-measurable sets are a more general definition. The last sentence of "Implications:" I said, "Defining how and where these non-measurable sets act on scale appears as a moving target and allows choice in how to define them." I used the term "assume" and "if we reject" a lot in "Implications".

This is when I discussed non-measurable sets in prior work such as Vitali sets and the Banach-Tarski theorem which allow re-constructions of objects of one size to objects of another size based on these non-measurable sets. Vitali sets specifically relates these non-measurable sets to reals.

The point I tried to make was that the Banach-Tarski paradox could be resolved if these non-measurable sets had exactly the same character (were equivalent) to scale choice, meters to feet for instance. I hoped someone would know of previous work related to this.

I wonder what I need to clean up the presentation? Thanks.

10. Mar 29, 2008

### ramsey2879

In your presentation you should use the slope intercept form specific to your diagram

Lines $$l_1 \quad l_2$$ have the slope 0 and each have the form the form y = b where b is the y intercept.

The line that passes through the orgin and has a positive slope has the y intecept of zero and is of the form y = mx where 0 < m < infinity

You can't sum sets in this manner as points have no dimension. The only parts that could be summed are the line segments and the sum of the all line segments from line $$l_1$$ and $$l_2$$ would always be S and -1/2 S where S approaches infinity.

There is no such thing as a non-measurable line segment so I don't think your diagram gives any insight to the "Banach-Tarski paradox" which I dont care to go into.

Last edited: Mar 29, 2008
11. Mar 29, 2008

### my_wan

The only two line I described using y = mx + b where $$l_1$$ and $$l_2$$, not the bijection line. Since the slope m was defined as zero y = 0x + b, which is y = b. The bijection line was as yet undefined and was never given a slope because it changes as you increment through the 1 to 1 process. This definition was only a definition to setup a relationship between $$l_1$$ and $$l_2$$ so that later the bijection would have meaning. I therefore fail to see why the form $$y - y_1 = m(x - x_1)$$ has any superior descriptive value. The term (y = mx + b) defines what slope form I'm using.

The slope of the bijection line was never defined, nor was defining it of any descriptive value. Its only purpose is to define a unique corresponding point on $$l_2$$ for any unique point on $$l_1$$, or visa versa. Even so under the definition I give the slope of the bijection line is limited to $$0 < m < \pm 1$$. Any definition that uniquely defined a point on one line by the choice of a unique point on the other line would work just as well.

That really depends on what definitions of infinity are used.

That is true in the same sense that the that a/b is undefined when a and b are infinite. However, if;
a = $$\infty = \sum_{n=1}^\infty n$$
and
b = $$\infty = \sum_{n=1}^\infty 2n$$
then a/b = 1/2. It really depends on what definitions of infinity is used.

So let's look at S in the limit you took. To meet the conditions I specified look at the inverse of the limit as you stated it. That is: The sum of the all line segments from line $$l_1$$ and $$l_2$$ would always be S and S/2 as S approaches zero. Now the questions I was going through in "Implications" was how do we define the difference in a point S and the derivative of S as S approaches zero (an infinitesimal). We could maintain the difference is in defining one as a limit and the other as an absolute point but that leaves a disjoint bewteen the magnitude of a line segement and the points that define that line. If we reject that a point is an infinitesimal in the derivative sense then we must define these infinitesimals as a seperate set of non-measurable sets. I considered the possiblitity of defining this difference wrt irrationals such that an infinitesimal was defined as the difference between a real and an irrational, such as 1.99... and 2, and found problems with it. A problem you seemed to have not noticed when you objected.

Compare this to the way Vitali sets {V} are defined. Wiki does ok here.
So in essence Vitali sets attaches a non-measurable set (infinitesimal) to the reals such that V $$\cap$$ [x] has a cardinality of 1. It's nothing more than a formal way of defining a point as an infinitesimal, i.e., a derivative. You can read the rest of the Wiki article to see the proof that Vitali sets are non-measurable sets.

What is $$dS = \lim_{S \to 0}f(S) = 0$$ if not a non-measurable line segment?

I can understand the reticence to get into the "Banach-Tarski paradox". However, perhaps you could answer far more limited question. Taking the axiom of infinity how do you define the difference between a point S and S + dx in the limit of x $$\to$$ 0?

12. Mar 29, 2008

### ramsey2879

half of an rational number is another rational number while half of an irrational number is just another irrational number. Also both rational lengths and irrational lengths can be constructed on graph paper and both can be measured. As for limits both x and 2x go to zero as x goes to zero but at different speeds.

13. Mar 29, 2008

### my_wan

Yes, hadn't thought of that way but essentially leads to the same problem with using irrationals the way I suggested and rejected. It's time to put the irrational issue to rest since I never claimed it was viable to begin with.

The wiki article on Vitali sets used "mass" rather than "speed" but to the same effect. So by attaching a "speed" to a particular point we have for all intent and purposes defined it as a derivative (as x approaches 0) the same way a Vitali set is defined.

Quoting the wiki article again; http://en.wikipedia.org/wiki/Vitali_set
So what my conjecture ask is: Is this "speed" you've attached to the points on this line an intrinsic property of the points or simply a product of the chosen scale?

If this "speed" isn't defined by any type of set how can you define a line segment as a series of points, which by definition sums to 0?

14. Mar 30, 2008

### ramsey2879

No a line segment is not a series of points. A line segment is a series of infinitesimals, i.e. the difference between two points. For any infinitesimal line segment there is a rational number or irrational number that corresponds to that line segment even if a scientist does not have the money to refine his instruments enough to measure it. So I take issue with the wiki article to the extent that it refers to a line segment as a series of points.

15. Mar 30, 2008

### my_wan

Yes, quiet true in certain axiomatic systems, but an infinitesimal is defined to be a segment so small that it is impossible to physically measure even in principle. Inverse example below.

Here you assume an infinitesimal has a finite extension, which is counter to the definition under the axiom of infinity I invoked. Consider the summation of a = 1+2+...+n in the limit of n. It is obviously infinite, not just super large. That does not mean that its ratio to another infinite series, b = 2+4+...+n, is no longer valid. Likewise, defining an infinitesimal as infinitely small, not just super small, does not invalidate its ratio to another infinitesimal in the limit. Given the axiom of infinity an infinitesimal is not a finite interval, and no amount of money or equipment is going to measure it. Are you assuming a summation of an infinite series can't produce exact answers?

You can rightfully object yet what they said was that it was an "arbitrary subset of the real line". They did not say that the "speed" of a number was so, they said it was a "natural question". The very motivation of defining Vitali sets was so as not to distort this distinction you make between points and intervals, yet try and provide some continuity between the statements. This is why a singleton Vitali set is defined as $$V \cap [x]$$. In essence what you are objecting to is the axiom of choice. What the Vitali theorem did, using the existence theorem and AC, is to prove that an infinitesimal is not Lebesgue measurable.

1) You've maintained that an infinitesimal is finite.
(Breaks the by definition inverse relation to infinite, i.e., infini (te) simal - infinite small and violates the Vitali theorem.)
2) You've maintained that an point is not finite.
3) You've maintained that a line segment cannot be defined, even in principle, by the points on it.

Your position is not that unreasonable yet what axioms are you denying as a result?
1) The axiom of infinity. (Which I invoked.)
2) The axiom of choice.
This also requires you to take the concept of a point as a purely imaginary construct, else you must take the axiom of infinity and reject AC.

If indeed you are willing to reject these axioms there is one question you have yet to answer.

1) A set of points are an arbitrary subset of an interval but cannot define the interval.
Therefore how do you define the set that is missing in our set of points to define an interval? You suggest an interval. Yet this interval, like the interval it defines, has an arbitrary set of points that can be defined on it. So we are back the original question of what set is missing that is not defined by these points to define that interval. In other words a set member that is defined by an interval cannot be considered a singleton.

So now I will ask the question this way. If a set of points on an interval can't define the interval, define a singleton set member, of any kind, that completes the definition of an interval.

Neither points nor other intervals have been deemed sufficient based on your previously stated position.

<NOTE>
I understand your position and being an axiomatic system I'm can't argue you are wrong. What is at issue here is the axiom of choice and the concomitant paradoxical decompositions of a unit interval, i.e., mathematical consistency wrt AC. The Banach–Tarski theorem was in fact modeled on the earlier work by Giuseppe Vitali dealing with interval decompositions. These theorems were proved within the confines of the axioms invoked. If we reject AC then there is nothing left to argue. Otherwise the above question stands as an important open issue of number theory. The one thing you can't do is maintain your position without rejecting AC. If you state straight out, "I'm rejecting AC", then you are correct. There is nothing left to argue. It simply lacks relevance within the axioms I have invoked.
</NOTE>

16. Mar 30, 2008

### ramsey2879

I do not reject the axiom of choice in that one can define an infinitesimal to be as small or say om the order of thousand times smaller than the diameter of the smallest known particle in the physical world in the same mannor as creating a smaller length standard by dividing a meter by a power of ten to define a millimeter but by using a higher power of ten. Mathematically it is possible to describe the width of any line segment that can be observed.

17. Mar 31, 2008

### my_wan

So what you are objecting to is the axiom of infinity?

Here's the problem: What Vitali's theorem proves, given AC, is that measurable sets cannot be the collection of all subsets if we hope to have a measure that returns the length of an open interval.
http://planetmath.org/encyclopedia/MeasurableSpace.html

Vitali theorem: There exists a set $$V\subset [0,1]$$ which is not Lebesgue measurable.
http://planetmath.org/encyclopedia/VitalisTheorem.html

Here is a proof of Vitali's theorem.
http://planetmath.org/?op=getobj&from=objects&id=4467

So no, you can't maintain this concept of measurability even in principle, much less physically. At least not without rejecting AC.

18. Mar 31, 2008

### ramsey2879

The only conflict in these two positions is that I accept the fact that one much chose a basic length from which to measure and can only give as length to the nearest unit of this measure. So one can not measure exactly that which is not in the equivalence class of the basic unit, but of course one can measure it the the nearest length that is in the equivalence class. Of course it is not possible to measure any length in any other way.

Last edited: Mar 31, 2008
19. Mar 31, 2008

### my_wan

Ok, that tells me a bit more of your thinking about it. Mostly I can't disagree here except to say that there are many equivalence classes. You seem to have neatly placed infinitesimals in the equivalence class of measurable in spite of the OP opening by invoking the axiom of infinity.

So let's look define some equivalence classes.
A ~ Lebesgue measurable
B ~ Infinite

We have 3 sets of objects.
x - Intervals
y - Infinitesimals
z - Points

Equivalence class membership.
(x, y) ~ A
(y, z) ~ B

Now where we seem to diverge is wrt placing y in the equivalence class of B wrt an infinite class. A Lebesgue measure of $$\infty$$ is well known to be possible. This does not break the rule that one cannot measure exactly that which is not in the equivalence class. For instance, one cannot measure $$dx$$ wrt a finite line segment. One can only measure it wrt another derivative, i.e., an entity of the same equivalence class (infinitesimal).

Grouping numbers into equivalence classes does not get you out of the quandary wrt the Vitali theorem and Lebesgue measure. So I'll ask the same question again another way. If a set of points on an interval can't define the interval, define a conjugacy class that completes the definition of an interval.

In the above post you have apparently admitted that this is impossible. Yet an interval is by definition is a superset of subsets. You said, "one much choose a basic length from which to measure". In what way does this differ from the conjecture I stated?

20. Apr 1, 2008

### ramsey2879

Well if you insist that (y, z) ~ B i.e. are in the same class then I understand your problem because the concept of a point like the concept of $$\infty$$ are just concepts and can't be used to measure anything that is concrete.

You ask me to deal with your conclusions so from my perspective I probably would reach different conclusions:

Implications:
"1) Any finite one to one mapping via origin bijection indicates that $$l_2/l_1 = 1$$, yet in the limit must equal 1/2 in this case, and varies generally."
As you noted the infintesimals must have a ratio of 1/2 so $$l_2 = \infty$$ while $$l_1 = 2\infty$$ This does not violate the concept of infinity.

"2) Implication 1) indicates the set of infinitesimals not contained in the subset R, which are neither measurable nor countable, must have a total magnitude (summation) equal to the total magnitude of the subset R defined in l."
True since $$\infty = 2\infty$$

"3) We can maintain that implication 2) is not due to a summation of infinitesimals not contained in the subset R but that indicates that 1 to 1 correspondence is invalid. Vitali sets specifically defines this as a property of reals."
I don't understand you at this point.

"4) We could assume the difference in $$l_1$$ and $$l_2$$ are a result of irrational numbers. This implies that the ratio of irrational numbers to real numbers can take any value, finite and otherwise, i.e., the irrational numbers defined on any line segment can sum to any value we choose simply by choice of scale."
5) Scale freedom (coordinate independence) is a fundamental principle and we could assume that the ratio of reals to irrationals has some actual value that changes due to scale. However, if this was the case, in principle the same bijection effects should apply to the irrationals.
6) If we reject this scale choice as a property of irrationals then we must assume a set of points not defined by any definable number must contain this property of scale freedom, i.e., not Lebesgue measurable. Assuming this is a property of irrationals means that the axiom of choice is the only assumption required to say 1.99... = 2. Defining how and where these non-measurable sets act on scale appears as a moving target and allows choice in how to define them.
"
The only agreement is that any measure has its own equivalence class, and that is all integer multiples of that measure are in the equivalence class, i.e. 1/2 ~ 3/2 and if one chooses a base measure anything not in its equivalence class would not be measurable exactly.

Last edited: Apr 1, 2008