GEOMETRIC MEASURE THEORY TUOMAS ORPONEN ABSTRACT. These are the lecture notes for the course Geometric measure theory, given at the University of Helsinki in fall semester 2018. The presentation is largely based on the books of Falconer [6] and Mattila [16, 18]. CONTENTS 1. Some discrete problems2 1.1. Applying the elementary incidence bound3 2. Dimension5 2.1. Hausdorff measures and Hausdorff dimension6 2.2. Some general measure theory8 2.3. Continuous analogues of Questions1,2 and3 10 3. The mass distribution principle and Frostman’s lemma 10 3.1. Frostman’s lemma 12 3.2. Weak convergence and compactness of measures 12 3.3. Proof of Frostman’s lemma 15 3.4. Hausdorff dimension of product sets 17 4. Riesz energies 18 4.1. Marstrand’s projection theorem: part I 20 5. Rectifiable sets and the Besicovitch projection theorem 23 5.1. A density lemma 23 5.2. Rectifiable and unrectifiable sets 24 5.3. Projections of rectifiable sets 25 5.4. Conical densities of purely 1-unrectifiable sets 26 5.5. Projections of unrectifiable sets 30 6. Fourier transforms of measures 36 6.1. Marstrand’s projection theorem: part II 39 6.2. Distance sets 40 6.3. Fourier dimension and spherical averages 43 7. Kakeya sets 46 7.1. The box dimension of planar Kakeya sets is 2 47 7.2. The pigeonhole principle, and the Hausdorff dimension of Kakeya sets 49 7.3. Kakeya sets in higher dimensions: the hairbrush argument 52 References 54 1 2 TUOMAS ORPONEN 1. SOME DISCRETE PROBLEMS This course is about studying the geometric properties of fractals using tools from measure theory (in addition to combinatorial and geometric arguments). What is a frac- tal? There is no rigorous definition, but usually people have something like Figure1 in mind. A fractal is typically uncountable, has zero Lebesgue measure, and is very "non- FIGURE 1. A fractal. smooth". Given a fractal, typical questions we will study are: how big is it, and how do various transformations affect its size? Since fractals can’t be (sensibly) measured by either counting or Lebesgue measure, we will need Hausdorff measures and Hausdorff dimension to quantify their size. These concepts are introduced in Section2. The questions we ask about fractals often have their origins in a field of combinatorics known as incidence geometry. So, before heading to measures, dimension and so on, we discuss three problems in incidence geometry, whose "fractal versions" will later occupy 2 us during most of the lectures. Let P ⊂ R be a set with cardinality jP j = n 2 N. 1 Question 1 (Projections). Let e 2 S (the unit circle), and let πe be the projection to the line 1 `e = span(e). How many directions e 2 S can there be such that jπe(P )j ≤ n=8? Question 2 (Distance sets). How many distinct distances does P span? In other words, find a lower bound for the size of the "distance set" ∆(P ) := fjp − qj : p; q 2 P g: 2 Question 3 (The Kakeya problem). Let m 2 N, and let L is a family of m distinct lines in R . Assume that jP \ `j ≥ m; ` 2 L: Find a lower bound for n = jP j. 2 Questions1 and3 are quite easy in R (but not in higher dimensions). To solve them, consider the following notion: d Definition 1.1 (Incidences). Let P ⊂ R be a set of points, and let L be a family of lines d in R . The incidences between P and L are the pairs I(P; L) := f(p; `) 2 P × L : p 2 `g: GEOMETRIC MEASURE THEORY 3 The fundamental question of incidence geometry is: how large can I(P; L) be (also when L is replaced by a collection of more complicated sets than lines)? The basic result 2 in R is the following sharp bound of Szemerédi and Trotter [20] from the 80’s: 2 2 Theorem 1.2. For any P ⊂ R and L a finite set of lines in R , we have 2=3 2=3 jI(P; L)j . jP j jLj + jP j + jLj: We will prove something weaker, but still useful: 2 2 Proposition 1.3. Let P ⊂ R be a finite set of points, and let L be a finite set of lines in R . Then jI(P; L)j ≤ 4 minfjP jjLj1=2 + jLj; jP j1=2jLj + jP jg: Note that the Szemerédi-Trotter theorem seems to "interpolate" between the two terms appearing in the proposition: it’s a much better bound when jP j ≈ jLj. Proof of Proposition 1.3. We only prove that jI(P; L)j ≤ 4(jP jjLj1=2 + jLj); since the other bound can be obtained with a similar argument, interchanging the roles of points and lines. We first estimate as follows, using the definition of I(P; L), and Cauchy-Schwarz: !1=2 X X jI(P; L)j = jP \ `j ≤ jLj1=2 jP \ `j2 : `2L `2L Then, we write jP \ `j2 = jf(p; q) 2 P × P : p; q 2 `gj, and exchange the order of summation. X X jP \ `j2 = jf` 2 L : p; q 2 `gj: `2L p;q2P Finally, we separate the "diagonal terms" where p = q: X X X jf` 2 L : p; q 2 `gj = jf` 2 L : p 2 `gj + jf` 2 L : p; q 2 `gj: p;q2P p2P p;q2P p6=q The first sum is simply jI(P; L)j again! For the second sum, note that jf` 2 L : p; q 2 `gj ≤ 1; 2 whenever p; q 2 R are distinct. Putting everything together, and using the basic inequal- ity (a + b)1=2 ≤ a1=2 + b1=2 yields 1=2 jI(P; L)j ≤ jLj1=2 jI(P; L)j + jP j2 ≤ jLj1=2jI(P; L)j1=2 + jLj1=2jP j: If the first term on the right is larger, then jI(P; L)j ≤ 2jLj1=2jI(P; L)j1=2, which gives jI(P; L)j ≤ 4jLj ≤ 4(jLj1=2jP j + jLj) 1=2 1=2 In the opposite case also jI(P; L)j ≤ 2jLj jP j ≤ 4(jLj jP j + jLj). 1.1. Applying the elementary incidence bound. We will now apply Proposition 1.3 to 2 Questions1 and3 in R . 4 TUOMAS ORPONEN 1.1.1. Solution to Question1. Let’s first look at Question1: consider the set of vectors 1 n B := fe 2 S : jπe(P )j ≤ 8 g; where n = jP j. By deleting at most half of the vectors in B, we may assume that no distinct vectors in B are parallel. For fun, let’s note that a "trivial" bound for jBj is jP j jBj ≤ ≤ jP j2; (1.4) 2 because for every e 2 B there exist distinct points p; q 2 P such that πe(p) = πe(q), and jP j there are only 2 possible choices of such fp; qg. Moreover, a fixed couple fp; qg can work for at most one vector in e 2 B, namely the one with e ? (p − q)=jp − qj. Proposition 1.3 will give something much better than (1.4). For each e 2 B, consider −1 the family of lines Le := fπe ftg : t 2 πe(P )g. Then jI(P; Le)j = jP j = n; (1.5) because every point in P lies on a unique line in Le. Then, consider further [ L := Le: e2B Note that the families Le are disjoint for different choices of e 2 B, because we assumed that B contains no pairs of parallel vectors. It now follows from (1.5) that jI(P; L)j = jBjn. Since jLej ≤ n=8 for all e 2 B, Proposition 1.3 gives jBjn1=2 jBjn jBjn = jI(P; L)j ≤ 4(jP jjLj1=2 + jLj) ≤ 4n · + : 8 2 Subtracting jBjn=2 from both sides and rearranging gives 1 jP j jfe 2 S : jπe(P )j ≤ 8 g ≤ 64jP j: It’s not hard to see that this estimate is sharp in the sense that jP j (on the right hand side) can’t be replaced by jP js for any s < 1. Probably 8 and 64 are not the best constants. 1.1.2. Solution to Question3. Next, we consider Question3. This is a straightforward application of Proposition 1.3 to the sets P; L. Since jP \ `j ≥ m for all ` 2 L, we have m2 ≤ jI(P; L)j ≤ 4(jP j1=2m + jP j); 2 now using the second inequality in Proposition 1.3. It follows that n = jP j & m , which is clearly optimal (up to multiplicative constants). Question3 is much harder in higher dimensions. A naive formulation is the following: d−1 d assume that L is a collection of m lines in R , each containing m points of P . How big is P ? One might first guess that d−1 d jP j & m · m = m ; 3 2 in analogy with the planar result. This already fails in R . In fact, if all the m lines in L 3 are all allowed to lie on a common plane V ⊂ R , then it’s possible to arrange them so, 3 and find a corresponding set P ⊂ V ⊂ R , such that jP \ `j ≥ m for all ` 2 L and jP j ≈ m5=2 m3: GEOMETRIC MEASURE THEORY 5 3 However, this is the only obstacle in R : Guth and Katz [9] have shown that if no plane 3 3 in R contains more than m lines, then jP j & m . In dimensions d ≥ 4, the question is open: one needs to assume that the lines are not concentrated on algebraic varieties of d low degree, but it’s not known if this is sufficient to guarantee jP j & n .