Shades of the Cauchy functional equation
and Segre functions

It follows that

Fact: Let f:R²®R be a discontinuous additive function. Then if J is any bounded rectangle in the plane, we again have

Fact: Let f:R²®R² be a discontinuous additive function. Then there are many other cool facts about f ^-1(checkerboards, stripes, half-planes, honeycombs...)

Speaking of planar sets...

In 1920, Sierpinski showed that CH implies that there exists a subset S of the plane with the property that S intersects all horizontal lines in a countable set and S intersects all vertical lines in a co-countable set.

Absurd. So CH is false.³

Sierpinski's set S is simple to "construct." First, take any well-ordering Ð of R and let

Thus a horizontal section, for any fixed y, has cardinality < c while the complements of vertical sections also have this property. If CH holds, they are countable.

Let sh ₁ X and sh ₂ X denote the 1- or 2-dimensional shade of a subset X in R or R², repectively.

Fact. For n = 1 or n = 2, if X Ì Rⁿ and |X| < c, then sh _n(X) = 0.

Simply put, the shade of a noncontinuum set must be zero. This is proved in [3].

Consequence #1: Sierpinski's set S has vertical shades all 1, horizontal shades all 0.

So if you throw darts in any vertical direction, you'll always hit it S. If you throw darts in any horizontal direction, you'll (almost surely) miss S.

Consequence #2: Then AC must also be false.

Consequence #3: Then my research program has just vanished and I must look for work elsewhere.

Consequence #4: Then AC must be okay after all.

That being the case, the following could be useful:

Theorem. Let L denote the set of lines in R² and let

At this point I described a practical application of such sets. You're in a bar or pub in some part of the world where AC is legal. You offer to play a game with one of the local denizens, wherein you each take turns throwing darts at a set F constructed in the plane of the floor. You each get the same number of throws, first at a set constructed by your opponent, then at one constructed by you. You each sit at right angles to each other, so you throw in the vertical direction, your opponent throws in the horizontal direction. (I didn't provide these details during the talk.) A player who hits F always collects a dollar (or beer, or whatever). The locals invariably construct simple measurable planar sets and the play is fair. But you, with your superior knowledge of AC, of course you construct Sierpinski's set S, so you win on every throw. Two problems emerge. One is that it is impossibly difficult to aim your throws in such exact vertical and horizontal directions. The second is that even if that problem were fixed, you must remember that just because the locals don't know AC doesn't mean that they are stupid. Since they never hit the set at all, they assume you are cheating, so they beat you up, steal your wallet, and throw you out of the joint.

Now here is the practical application of sets such as A in the above theorem. You construct your set so that all lines passing through the dartboard and a vicinity of your chair have a shade of, say, 53/100, while lines through your opponent's vicinity have shade 47/100. This solves the problem of having to aim in an exact direction, as you would for Sierpinski's set S, and also, your opponent doesn't really notice the slight advantage you have. Eventually, you win most of the throws. (You still might get beat up, though.) Here is a crude diagram of the idea.

Problem. Suppose that A is a subset of the plane whose linear shades are all equal to t. What is the planar shade of A (if it even exists)?

The "obvious answer" is t. But there is nothing obvious in the AC world, as the following indicates.

Fact: (Kharazishvili (Tbilisi), personal communication) A subset B of the plane exists for which

1)

|BÇL| < c for each line L Ì R²,

2)

l₂^*(BÇE) = l₂(E) for each Lebesgue measurable set E Ì R² (where l₂^* is the Lebesgue outer measure on R²),

3)

|BDg(B)| < c for all g in the group D₂ of all isometries on R².

This implies the existence of a D₂-invariant measure n (on the s-algebra generated by B and the Lebesgue measurable sets) which is concentrated on B.

So for any Lebesgue measurable set E Ì R²,

This means that every linear shade of B is zero, but the planar shade is not.

However, the set B doesn't have a "uniqueness property" in that one can (I think) also define, for the same B, a measure n¢ concentrated on R²\B.

At least we can do this much:

Theorem: For any t Î [0,1] there exists T Ì R² for which

(1)

sh _L(T) = t for each L Î L,

(2)

sh ₂(T) = t.

Such a T can be constructed as follows. Let

(a)

f(x y) = f(x)f(y),

(b)

f(x+y) = f(x)+f(y).

(See Kuczma's book [2] or the paper of Kestelman [1].)

The set T = f ^-1(D_t) does the trick. To prove it, use the fact that for all nonempty open sets U in R², the set f ^-1(U)ÇL is dense in every line L in C. This is used to prove property (1). The additive property (b) of f yields the two-dimensional shade (property (2)).             ¨

It is an open problem whether or not a set C Ì R² exists for which all linear shades exist and are equal to some fixed s₁, while the planar shade s₂ also exists but for which s₁ ¹ s₂ (as opposed to the set in the previous theorem, in which s₁ = s₂). It seems inconceivable to me that this can happen, but some experts in measure theory warn me that the one- and two-dimensional measure extensions we are dealing with here are fairly independent.

Bibliography

[1]

Kestelman, H., Automorphisms of the field of complex numbers, Proc. London Math. Soc., II. Ser., 53 (1951) 1-12.

[2]

M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities, Pa\'nstwowe Wydawnictwo Naukowe, 1985.

[3]

R. Mabry, Sets which are well-distributed and invariant relative to all isometry invariant total extensions of Lebesgue measure, Real Analysis Exchange, 16 (1991-1992) 425-459.