Dimensional problem

The measure problem is a problem in mathematics that is fundamental to measure theory . It raises the question of whether one can reasonably assign a measure of area, often called area , to every subset of the plane, that is to say every set composed of points on the plane in any way . Depending on what is supposed to be considered reasonable, the surprising answer is no . The corresponding question must be answered negatively not only in the two-dimensional level, but also in other dimensions.

The dimensional problem in the plane

Increasingly complex subsets of the plane and their areas

Even in elementary school mathematics one learns to assign numbers as their area to certain subsets of the plane, which are called areas . A rectangle is assigned the area “side times side”, a triangle is assigned “base side times height by 2”. Curvilinearly bounded areas such as circles ( times the radius to the square) are added later, and areas of area are considered under curves even later, and the methods for determining the area become increasingly complex. Can you go on and on and finally assign a number as area to each subset of the level in a reasonable way? ${\displaystyle \pi }$

The following demands will be made on a reasonable area measure that assigns a number as area to every set , i.e. every subset of the plane (the Greek letter μ is intended to remind of measure ): ${\displaystyle A\subset \mathbb {R} ^{2}}$ ${\displaystyle A}$ ${\displaystyle \mu (A)}$

Positivity : For every crowd is , where is also allowed. Negative numbers are not allowed as areas. ${\displaystyle A\subset {\mathbb {R} }^{2}}$ ${\displaystyle \mu (A)\geq 0}$ ${\displaystyle \infty }$

Congruence : If two sets are congruent , i.e. H. they emerge from one another by parallel displacement , mirroring or rotation , so they should have the same area measure, i.e. H. it should apply. ${\displaystyle A,B\subset \mathbb {R} ^{2}}$ ${\displaystyle \mu (A)=\mu (B)}$

Normalization : The unit square should have the area measure 1, i.e. H. . This rules out the simple but not very useful solution of setting all areas to 0. ${\displaystyle [0,1]^{2}\subset \mathbb {R} ^{2}}$ ${\displaystyle \mu \left([0,1]^{2}\right)=1}$

Completion of a circle by triangles

For the next property, let's remember how a circle can be scooped up by a series of ever smaller triangles. You begin by inscribing an equilateral triangle on the circle. Then you place an isosceles triangle on each side of the triangle , the tip of which lies on the circular line and whose base should not belong to it, and you get a hexagon inscribed on the circle. By repeating this, you get a dodecagon , then a 24-corner, and so on. You can see that the circle is made up of an infinite number of non-overlapping triangles. The area of ​​the circle is then calculated as the "infinite sum" of the ever smaller triangles. In order for such constructions to be successful, the following requirements must be made on the area measure:

${\displaystyle \sigma }$- Additivity : Is a sequence of non- point sets in pairs , i. H. for , as the infinite to association the series have a surface measure, d. H. in compact notation: if for everyone . ${\displaystyle A_{1},A_{2},A_{3},\ldots \subset \mathbb {R} ^{2}}$ ${\displaystyle A_{i}\cap A_{j}=\emptyset }$ ${\displaystyle i\neq j}$ ${\displaystyle A_{1}\cup A_{2}\cup A_{3}\cup \ldots }$ ${\displaystyle \mu (A_{1})+\mu (A_{2})+\mu (A_{3})+\ldots }$ ${\displaystyle \mu \left(\bigcup _{i=1}^{\infty }A_{i}\right)=\sum _{i=1}^{\infty }\mu (A_{i})}$ ${\displaystyle A_{i}\cap A_{j}=\emptyset }$ ${\displaystyle i\neq j}$

That in the term -additivity should indicate that countable infinite sums are being treated here. ${\displaystyle \sigma }$ ${\displaystyle \sigma }$

If one denotes the set of all subsets of as usual , then the measurement problem in its precise form reads: ${\displaystyle \mathbb {R} ^{2}}$ ${\displaystyle {\mathcal {P}}\left(\mathbb {R} ^{2}\right)}$

• Is there a mapping that fulfills the above-mentioned properties of positivity, congruence, normalization and additivity? ${\displaystyle \mu \colon {\mathcal {P}}\left(\mathbb {R} ^{2}\right)\rightarrow [0,\infty ]}$ ${\displaystyle \sigma }$

The dimensional problem in n dimensions

The problem of dimensions can easily be extended to -dimensional spaces; this includes the dimensions 1, 2 and 3 accessible to visualization, whereby one then has to do with length, area and volume measurements. The measurement problem in dimensions is: ${\displaystyle n}$ ${\displaystyle n}$

Is there a figure with the following properties: ${\displaystyle \mu _{n}:{\mathcal {P}}\left(\mathbb {R} ^{n}\right)\rightarrow [0,\infty ]}$

• Positivity : for everyone (this condition is actually already in the specification of the image area of the figure), ${\displaystyle \mu _{n}(A)\geq 0}$ ${\displaystyle A\subset \mathbb {R} ^{n}}$
• Congruence : if A and B are congruent, ${\displaystyle \mu _{n}(A)=\mu _{n}(B)}$
• Normalized awareness : , ${\displaystyle \mu _{n}\left([0,1]^{n}\right)=1}$
• ${\displaystyle \sigma }$-Additivity : if for everyone  ? ${\displaystyle \mu _{n}\left(\bigcup _{i=1}^{\infty }A_{i}\right)=\sum _{i=1}^{\infty }\mu _{n}\left(A_{i}\right)}$ ${\displaystyle A_{i}\cap A_{j}=\emptyset }$ ${\displaystyle i\neq j}$

One could even make further obvious demands on a measure function:

Empty set : , ${\displaystyle \mu _{n}(\emptyset )=0}$

One point set : for all points , ${\displaystyle \mu _{n}\left(\{x\}\right)=0}$ ${\displaystyle x\in \mathbb {R} ^{n}}$

Additivity : if , or ${\displaystyle \mu _{n}(A_{1}\cup A_{2})=\mu _{n}(A_{1})+\mu _{n}(A_{2})}$ ${\displaystyle A_{1}\cap A_{2}=\emptyset }$

Monotony : if . ${\displaystyle \mu _{n}(A)\leq \mu _{n}(B)}$ ${\displaystyle A\subset B}$

These properties can easily be recognized as simple consequences of the properties formulated at the beginning, so they do not represent any further requirements.

Unsolvability of the measurement problem

Here, by means of an idea that goes back to the Italian mathematician Giuseppe Vitali , it is shown that the dimensional problem cannot be solved for the one-dimensional case . To do this, you construct a set , the so-called Vitali set , with the following properties: ${\displaystyle {\mathbb {R} }^{1}={\mathbb {R} }}$ ${\displaystyle V\subset [0,1]}$

1. If and are different rational numbers , so is . Where is the amount shifted. ${\displaystyle p}$ ${\displaystyle q}$ ${\displaystyle (p+V)\cap (q+V)=\emptyset }$ ${\displaystyle p+V}$ ${\displaystyle p}$
2. If the sequence is a counting of the rational numbers contained in the interval , then is . ${\displaystyle (p_{n})_{n\in {\mathbb {N} }}}$ ${\displaystyle [-1,1]}$ ${\displaystyle \textstyle [0,1]\subset \bigcup _{n=1}^{\infty }p_{n}+V}$

From this it follows easily that the measurement problem cannot be solved. If there were a solution, because of the second property and because each one lies in the interval : ${\displaystyle \mu }$ ${\displaystyle p_{n}+V}$ ${\displaystyle [-1,2]}$

${\displaystyle [0,1]\subset \bigcup _{n=1}^{\infty }p_{n}+V\subset [-1,2].}$

As you consider light that must be. Hence the monotony results ${\displaystyle [-1,2]=[-1,0)\cup [0,1]\cup (1,2]}$ ${\displaystyle \mu ([-1,2])=3}$

${\displaystyle 1\leq \mu \left(\bigcup _{n=1}^{\infty }p_{n}+V\right)\leq 3.}$

According to the first property, the sets are pairwise disjoint, so that from the additivity ${\displaystyle p_{n}+V}$ ${\displaystyle \sigma }$

${\displaystyle 1\leq \sum _{n=1}^{\infty }\mu (p_{n}+V)\leq 3}$

follows. Now the sets are all congruent, because they emerge from one another through shifts. Hence the numbers are all the same. Then the infinite sum can only be 0 or , in any case it cannot be between 1 and 3. This contradiction shows that the measurement problem cannot be solved. ${\displaystyle p_{n}+V}$ ${\displaystyle \mu (p_{n}+V)}$ ${\displaystyle \infty }$

The task remains to specify the quantity : is a subgroup . Each minor class of contains elements from . With the help of the axiom of choice , we select one such element from each secondary class of and combine them to form a set . Then in included, and you think about it easy to see that the two above-mentioned properties. The first property applies because each secondary class contains exactly one element. To prove the second property, let us say anything. This must be in some minor class, because the union is across all minor classes . Since every minor class contains an element, there is one that is in the same minor class as , so the following applies to this . Since , and both are between 0 and 1, is also d. H. there is a with . Then follows what ends the proof. ${\displaystyle V}$ ${\displaystyle \mathbb {Q} \subset \mathbb {R} }$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle [0,1]}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle V}$ ${\displaystyle V}$ ${\displaystyle [0,1]}$ ${\displaystyle V}$ ${\displaystyle V}$ ${\displaystyle x\in [0,1]}$ ${\displaystyle x}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle V}$ ${\displaystyle v\in V}$ ${\displaystyle x}$ ${\displaystyle x-v\in \mathbb {Q} }$ ${\displaystyle x}$ ${\displaystyle v}$ ${\displaystyle x-v\in [-1,1]}$ ${\displaystyle m}$ ${\displaystyle x-v\,=\,p_{m}}$ ${\displaystyle \textstyle x=p_{m}+v\in p_{m}+V\subset \bigcup _{n=1}^{\infty }p_{n}+V}$

Thus the unsolvability of the measurement problem is shown to be complete. This immediately results in the unsolvability in higher dimensions, because if a solution of the measurement problem were for the dimension , then it would be through ${\displaystyle {\mathbb {R} }^{1}}$ ${\displaystyle \mu _{n}}$ ${\displaystyle n}$

${\displaystyle \mu _{1}(A):=\mu _{n}(A\times [0,1]^{n-1})}$

given a solution to the one-dimensional case, the non-existence of which we have just convinced ourselves.

Historical remarks

Mathematicians of the bygone days were rather careless with terms such as area or volume. It was only through Cantor's set theory and the associated stricter formalization of mathematics that such problems could even be recognized and formulated. The oldest formulation of the measurement problem goes back to Henri Léon Lebesgue . In his dissertation, submitted in 1902, he points out that it cannot be solved in general, but only for a certain class of sets, which he called measurable . Already in 1904 Giuseppe Vitali was able to show that the measurement problem was unsolvable.

Measuring the size of a set (length, area, volume) is one of the most important tasks in mathematics, which is of fundamental importance not only for applications but also for many mathematical theories. One way out of this dilemma could be to weaken the demands on measure. This has Felix Hausdorff done in 1914 by the replaced additivity by the weaker requirement of additivity. One then speaks of a content rather than a measure. It is clear that Vitali's argument for content collapses, and Felix Hausdorff was actually able to show that there is content in dimensions 1 and 2, but there is no longer any content either (see Banach-Tarski paradox ). ${\displaystyle \sigma }$ ${\displaystyle n\geq 3}$

Another way out has proven to be much more successful. The stronger technical requirement of -additivity, which is important for many proofs, is retained and the range of quantities to which a measure is assigned is reduced, as Lebesgue already did in his doctoral thesis. This resulted in the measurement theory and Lebesgue's integration theory . Felix Hausdorff wrote in 1914 in his book Basics of Set Theory : ${\displaystyle \sigma }$

The theory of content has developed in two stages, which are best characterized by two laws for additive behavior. The concept of content of the older level, based on the approaches of G. Cantor and H. Hankel and completed by G. Peano and C. Jordan , satisfies the equation

${\displaystyle f(A+B)\,=\,f(A)+f(B)}$

for two (or finitely many) pairwise foreign sets; the concept of content of the newer level, which we owe to E. Borel and H. Lebesgue , satisfies the equation

${\displaystyle f(A+B+C+\cdots )=f(A)+f(B)+f(C)+\cdots }$

for a finite or countable set of summands. The older level corresponds to the Riemannian, the new to the Lebesgue integral term. The transition from the finite to the countable can be described as one of the greatest advances in mathematics ...

So you limit yourself to the so-called Lebesgue measurable quantities. Vitali's argument then shows that the set V given above cannot be Lebesgue measurable, which is also known as Vitali's theorem . The method of obtaining this set V is not constructive since the axiom of choice was used. The question of whether the use of the axiom of choice (or a weaker variant of this axiom) is mandatory was answered with yes by Robert M. Solovay in 1970 , when he gave a model of set theory in which the axiom of choice does not apply and in which every subset of is measurable. ${\displaystyle \mathbb {R} }$

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• Felix Hausdorff : Fundamentals of set theory . Veit & Co, Leipzig 1914 (numerous reprints).
• Donald L. Cohn: Measure Theory. Birkhäuser, Boston MA et al. 1980, ISBN 3-7643-3003-1 .
• Robert M. Solovay : A model of set theory in which every set of reals is Lebesgue measurable. In: Annals of Mathematics . Second Series, Vol. 92, No. 1, 1970, pp. 1-56.