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Copyright   James R Meyer    2012 - 2024 https://www.jamesrmeyer.com

Lebesgue Measure Cranks

Page last updated 4 Nov 2024

 

Crank 1

A reader sent me an email, claiming that he could prove that my analysis of Lebesgue measure at An example of Lebesgue measure could not be correct - by claiming that he could demonstrate a specific example of an irrational number in the set B that could not be an endpoint of any complete interval of the set A. His claimed proof of this is the following (note that an interval of A is called a “complete interval” of A if it is not a sub-interval of any interval of A apart from itself):

 

 

Let  φ = √5 − 1 2 ≈ 0.618…   (incidentally φ is the inverse of the golden ratio).

We will prove that φ is neither in nor at the boundary of any complete interval.

 

Lemma. For all p, q ∈ N, if 0 < p < q we have |φ − p⁄q| > 1⁄3q2 .

 

Proof. Let

f (x) = x2 + x − 1 = (x − φ)(x + φ + 1)

Then we have:

Equation 1

Since f  has no rational roots f (p⁄q) ≠ 0. Therefore:

Equation 2

Meanwhile, we also have:

Equation 3

Thus:

Equation 4

Now we can prove that φ is neither in nor at the boundary of any complete interval.

 

For any complete interval, among all rational numbers it contain, there must be one that appears before other rationals in the given ordering.

 

Let that rational be p⁄q and let p⁄q be the nth rational in the ordering. Then we have n ≥ q − 1.

 

Then, the length of the complete interval containing p⁄q is at most

Equation 5

However, if q ≥ 3 then 9 × 10q - 2 > 3q2, so we have

Equation 6

and if q = 2 we can verify directly that

Equation 7

Thus, φ is neither in nor at the boundary of the complete interval containing p⁄q.

 

 

It's immediately obvious that there must be something wrong with the above “proof”, since the result is saying that the irrational φ is an interior point of a non-degenerate interval within which there cannot be any interval of the set A (note that every interval of A is necessarily a sub-interval of a complete interval of A). But that is impossible, since in every non-degenerate interval there exist infinitely many rationals, each of which is the midpoint of an interval of A.

 

But there’s an even simpler way of demonstrating that it cannot be correct - simply choose an enumeration of rationals that has 62⁄100 as its first rational, then the first interval 1⁄10 wide obviously includes 0.618… since its endpoints are 0.57 and 0.67.

 

The reason for the contradiction lies in the very first line of the “proof” - the author provides no reason why the function x2 + x − 1 might have anything to do with the intervals of the set A. This is simply a function that happens to intersect the x-axis at one point between 0 and 1. And there are infinitely many functions where the curve of the function crosses the x-axis at one point between 0 and 1. We can plot this particular function, a parabola, over the interval 0 to 1:

Graph of function

The author has simply selected this function without providing any logical rationale, creating a spurious relationship between the value of that specific function (the y-distance from the x-axis) at the point p⁄q and the value that is the difference between p⁄q and φ along the x-axis - a relationship has no correlation whatsoever with the determination of the points of the sets A and B in the case discussed at An example of Lebesgue measure.

 

In summary, the above “proof” is asinine crankery, not mathematics. Manipulation of mathematical symbols is pointless if sequences of symbols are haphazardly introduced without any underlying logical rationale.

Crank 2

The author of the following “proof” claimed that my analysis of Lebesgue measure at An example of Lebesgue measure could not be correct. He attempted to post it as a comment, but it was devoid of any sort of reasonable formatting, making it extremely difficult to read, and for that reason I moved it from the comments section to another page. I have now formatted his “proof” here so that anyone can easily read it (you can see the unformatted original at Moved comments for this site). The author claims that he can construct an irrational number that does not lie in any complete interval of A, nor is an endpoint of such a complete interval; his “proof” of this is the following (note that an interval of A is called a “complete interval” of A if it is not a sub-interval of any interval of A apart from itself):

 

 

Let us have an enumeration of the rational numbers in (0,1), q1, q2, …, qn, … . The set A is defined by the construction laid above of taking a 1⁄10n open interval around qn and taking the union of all such intervals. We label each open interval as Bn = I(qn, 1⁄10n) (i.e:the open interval around qn of radius 1⁄10n).

 

As Mr. Meyer nicely pointed out, all the complete intervals can be enumerated. So let us enumerate them as In with endpoints an and bn, where an  ≤  bn. Note that In is a subset of A, so every element of In must be an element of Bk for some k.

 

Aim:

to construct a number r in (0,1) which is neither equal to an nor equal to bn, nor belongs to Bn, for any n.

 

Implication:

if such an r exists, then it follows that this number does not belong to any complete interval In, nor is one of the endpoints an or bn, and thus, the measure of the set B is being undercounted in Mr. Meyer’s argument.

 

Construction:

Expand an, bn, and qn in decimal and take the nth digit of each. Call the nth digit xn, yn and zn of an, bn, and qn respectively. There are 10 possible decimal digit values, and if we remove the digits xn , yn , zn , zn-1 , zn+1 , 0 and 9 (not all of which may be valid digits) there are at least 3 other digit values left. So pick any one of those values, say wn , that does not equal xn , yn , zn , zn-1 , zn+1 , 0 or 9

 

Now, we create the number r = 0.w1w2…wn… , i.e: the nth digit is wn. We show that r is the number we need.

 

Clearly, r does not equal an or bn, because the nth digit does not match (by construction).

 

Let us show that r does not belong to any of the intervals Bk = I(qk, 1⁄10k). That will imply that r does not belong to A, and hence it does not belong to any complete interval.

 

If possible, let |r - qk| < 1⁄10k for some k. We expand qk in its decimal expansion, qk = 0.t1t2t3…. Note that, in our previous notation, tk = zk (the digit value).

 

Let j be the first digit where r and qk differ. So,

| wj - tj 10 j + wj+1 - tj+1 10 j+1 + … |  <  1 10k

 

Case 1:   Suppose j = k.

 

By the reverse triangle inequality |a| - |b| ≤ |a-b|,

|wk - tk| 10 j  −  d  <  1 10k

where d = | wk+1 - tk+1 10k+1 + wk+2 - tk+2 10k+2 + … |

Note that d  ≤  |wk+1 - tk+1| 10k+1 + |wk+2 - tj+1| 10k+2 + …

Since we are dealing with decimal digits

d  ≤  9 10k+1 + 9 10k+2 + 9 10k+3 …  ≤  1 10k

 

Hence,

|wk - tk| 10k   <   d + 1 10k   ≤   2 10k

implies |wk - tk|  <  2. But this is absurd because tk = zk, and wk was a specifically chosen digit so that it would not equal zk, zk-1, zk+1.

 

Case 2:   j ≠ k.

 

But we know that r and qk differ at the kth digit anyway, so we must have j < k.

 

So we simplify the beginning inequality,

| wj - tj 10 j + wj+1 - tj+1 10 j+1 + … |  <  1 10k

using the reverse triangle inequality as in Case 1, but slightly differently,

|wj - tj| 10 j − |wj+1 - tj+1| 10 j+1  −  …  −  |wk-1 - tk-1| 10k-1 − |wk - tk| 10k  −  d   <  1 10k

where

d = | wk+1 - tk+1 10k+1 + wk+2 - tk+2 10k+2 + … |

So

|wj - tj| 10 j   <   |wj+1 - tj+1| 10 j+1  +  …  +  |wk-1 - tk-1| 10k-1 + |wk - tk| 10k  +  d  +  1 10k

 

Again d  <  1 10k  (as in Case 1). Also, |wk - tk| 10k   ≤   8 10k , because wk does not equal 9 or 0 by construction, and so the difference of the digits |wk - tk| has to be less than 9.

 

On the other hand,

|wj+1 - tk+1| 10 j+1   ≤   9 10 j+1 ,

and

|wj+2 - tk+2| 10 j+2   ≤   9 10 j+2 ,

and so on.

 

So making the substitutions, we get

|wj - tj| 10 j   <   9 10 j+1 + 9 10 j+2 + … + 9 10k-1 + 8 10k + 1 10k + 1 10k

implies

|wj - tj| 10 j   <   9 10 j+1 + 9 10 j+2 + … + 9 10k-1 + 10 10k

implies (telescoping the sum from the right hand side)

|wj - tj| 10 j   <   1 10k

implies

|wj - tj|   <   1

 

But |wj -tj| ≥ 1 because they are distinct digits. Contradiction.

 

Clearly, r is not in Bk for any k. So we have found an irrational number that is neither the end point of nor contained in a complete interval.

 

 

The author’s conclusion leads directly to the contradiction that there must be two endpoints of complete intervals of A, between which there are no other points of A, but which includes this purported irrational. This is a direct contradiction since between any two reals there exists infinitely many rationals, and by definition, every rational is covered by an interval that is an interval of A.

 

But anyway, let’s examine the argument to see how the fallacy arises. The author selects a rational number with one decimal place, making sure that that value is not within 1⁄10 of the first rational in the enumeration of rationals, then selects another rational with that decimal place and adds another decimal place such that the number value is not within 1⁄100 of the second enumeration of the rationals, then selects another rational with the same two decimal places and adds a further third decimal place such that the number value is not within 1⁄1000 of the third enumeration of the rationals. He continues selecting rationals in this way.

 

He finally declares that he has constructed an irrational that is not within any 1⁄10n of any rational.

 

But he completely evades the question of when and how his sequence of rationals magically transforms into an irrational. There's no point in filling a page with mathematical manipulations if you’re simply going to dodge the fundamental underlying problem.

 

It’s rather ironic that the argument used is basically the same argument that Cantor uses in his 1874 proof, where, if there is a series of nested intervals (each inside another interval) then while the nesting is infinite and never actually reaches a stage where there remains only one point, there must be a point that is not the endpoint of any of the nested intervals. The difference is that in his paper, Cantor correctly notes that this point is the limit value both of the sequence of left endpoints, and of the sequence of right endpoints.

 

The reality is that infinitely nested intervals that are all defined as non-degenerate do not somehow magically “nest down” to a single point, a single point is only obtained by applying a limit. And the fact is that regardless of how large n becomes, there is no finite n at which an irrational number somehow emerges from the author’s description. But the author asks us to assume that his description, which is only in terms of finite n, without the inclusion of any limiting condition, results in a well-defined set - but it is a result that has irrational points are not in the set A, but at the same time, there must somehow be rational endpoints of the defined open intervals of A that are not in the set A - an absurd contradiction.

 

The author’s description only gives, for any chosen n, a rational number that has n decimal digits, and his calculations are all based around finite n - but an irrational has limitlessly many digits. There’s no point in pretending that calculations involving finite n are relevant when we are working with real numbers which have no terminal finite n, and whose value is the limiting value as that n continues to increase. As I explain elsewhere, if we suppose that there could be a sum of infinitely many digits, since every fraction 1⁄10n is inherently connected to the initial 1⁄10 by the finite factor n, then there would have to be at least one 1⁄10n where n is greater than any finite number - which is absurd. See Sums of infinitely many fractions: 1 and Sums of infinitely many fractions: 2 for a detailed explanation of why the notion of infinite sums without the application of a limit is contradictory. On the other hand, what we can logically assert is that, given the author’s description, we can define an irrational number that is the limit of the summation of the fractions that he describes.

 

The irony here is that by applying a limit to the author’s description, we obtain irrational limit points that are the endpoints of complete intervals of the set A as described in An example of Lebesgue measure and where a limit condition is applied. Rather than resulting in an absurdity such as in the author’s crank mathematics, the correct application of limits gives a simple and complete resolution of all the contradictions generated by the naive description of the set A.

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