Some properties of Lebesgue measure

Let \(\lambda\) be the Lebesgue measure on \(\mathbb{R}\). Then we have

1. \(\lambda(\{x\})=0\) for any \(x \in \mathbb{R}\),

2. \(\mathbb{Q}\) has Lebesgue measure \(0\). In fact, \(\lambda(U)=0\) for any countable set \(U \subseteq \mathbb{R}\),

3. If \(U\) is open in \(\mathbb{R}\), then \(\lambda(U)>0\),

4. If \(E\) is a null set in \(\mathbb{R}\) (i.e. \(\lambda(E)=0\)) then \(E^c\) is dense in \(\mathbb{R}\).

   Proof. For any open interval \(I \subseteq \mathbb{R}\) we have \(\lambda(I)>0\). Since \(\lambda(E)=0\) this means that \(E\) cannot contain any open interval. Thus \(E^c \cap I \neq \varnothing\) for any interval \(I \subseteq \mathbb{R}\) but this means that \(E^c\) is dense in \(\mathbb{R}.\)

Nowhere dense set with positive Lebesgue measure

1. 5. Let \(\mathbb{Q} \cap [0,1]=\{r_j: j \in \mathbb{N}\}\) and given \(\varepsilon>0\) let \(I_j\) be the open interval of length \(\varepsilon2^{-j}\), which is centered at \(r_j\). Then the set \[U=(0,1) \cap \bigcup_{j=1}^{\infty}I_j\] is open and dense in \(\mathbb{R}\), since \(U \supseteq \mathbb{Q} \cap (0,1)\). Thus \(U\) is topologically large but small in the sense of measure, since \[\lambda(U) \leq \sum_{j=1}^{\infty}\lambda(I_j) \leq \sum_{j=1}^{\infty}\frac{\varepsilon}{2^j}=\varepsilon.\]

2. 6. If \(K=[0,1] \setminus U\) then \(K\) is closed and nowhere dense thus topologically small but large in the sense of measure since \[\lambda(K)=1-\lambda(U) \geq 1-\varepsilon.\]

Accumulation point, isolated point, perfect set

There exists a perfect set in \(\mathbb{R}\) which contains no segment.

• Let \(C_0=[0,1]\). Given \(C_n\) that consists of \(2^n\) disjoint closed intervals each of length \(3^{-n}\) take each of these intervals and delete the open middle third to produce two closed intervals each of length \(3^{-n-1}\).

• Take \(C_{n+1}\) to be the union of \(2^{n+1}\) closed intervals so formed and continue.

Cantor set

• Each \(C_0 \supseteq C_1 \supseteq C_2 \supseteq \ldots\) is closed and bounded thus compact, and the family \((C_n)_{n \in \mathbb{N}}\) has finite intersection property thus the Cantor set is compact and \(\mathcal{C} \neq \varnothing\) .

Properties of the Cantor set

• Every segment \((\alpha,\beta)\) contains a segment of the form (*) if \(m\) is sufficiently large, since the set \[\left\{\frac{\ell}{3^m}: m \in \mathbb{N} \text{ and }0 \leq \ell \leq 3^{m}-1\right\}\] is dense in \([0,1]\). Thus \(\mathcal{C}\) contains no segment \((\alpha,\beta)\). This also shows that \({\rm int\;}\mathcal C=\varnothing\).

• To prove that \(\mathcal{C}\) is perfect it is enough to show that \(\mathcal{C}\) contains no isolated points. Let \(x \in \mathcal{C}\) and let \(I_n\) be the unique interval from \(C_n\) which contains \(x \in I_n\). Let \(x_n\) be the endpoint of \(I_n\) such that \(x \neq x_n\). It follows from the construction of \(\mathcal{C}\) that \(x_n \in \mathcal{C}\). Hence \(x\) is a limit point of \(\mathcal{C}\) thus \(\mathcal{C}\) is perfect.

Cantor set has Lebesgue measure zero

• We have \[\lambda(\mathcal C)=0.\]

  Indeed, by the construction note that \[\mathcal{C}=\bigcap_{n=0}^{\infty}C_n\] and \[\lambda(C_n)=\bigg(\frac{2}{3}\bigg)^n \quad \text{ for any } \quad n\in\mathbb N.\] By the continuity of \(\lambda\), since \(C_0 \supseteq C_1 \supseteq C_2 \supseteq \ldots\) and \(\lambda(C_0)=1\), we have \[\lambda(\mathcal C)=\lim_{n\to \infty}\lambda(C_n)=\lim_{n\to \infty}\bigg(\frac{2}{3}\bigg)^n=0.\]

More about Cantor set

• Each component of \(C_n\) can be described as the set

  \[C_n=\left\{\sum_{n=1}^\infty \frac{\varepsilon_j}{3^j}\;:\; \varepsilon_j \in \{0,1,2\} \text{ and }\varepsilon_j \neq 1 \text{ for }1 \leq j \leq n\right\}.\].

• Consequently,

  \[{\color{teal}\mathcal{C}=\left\{\sum_{n=1}^\infty \frac{\varepsilon_j}{3^j}\;:\; \varepsilon_j \in \{0,2\} \right\}.}\].

Fact

Proof. Take \(\varepsilon=(\varepsilon_j)_{j \in \mathbb{N}}\), \(\delta=(\delta_j)_{j \in \mathbb{N}}\) with \(\varepsilon_j,\delta_j \in \{0,2\}\) such that \(\varepsilon \neq \delta\). Let \(N=\min\{j \in \mathbb{N}\;:\; \varepsilon_j \neq \delta_j\}\) and assume \(0=\varepsilon_N<\delta_N=2\). Then \[\begin{align*} \sum_{j=1}^{\infty}\frac{\varepsilon_j}{3^j}&= \sum_{j=1}^{N-1}\frac{\varepsilon_j}{3^j}+\sum_{j=N+1}^{\infty}\frac{\varepsilon_j}{3^j} \leq \sum_{j=1}^{N-1}\frac{\delta_j}{3^j}+\frac{2}{3^{N+1}}\sum_{j=0}^{\infty}\frac{1}{3^j} \\&\leq \sum_{j=1}^{N-1}\frac{\delta_j}{3^j}+\frac{2}{3^{N+1}}\underbrace{\frac{1}{1-\frac{1}{3}}}_{{\color{red}\frac{3}{2}}} =\sum_{j=1}^{N-1}\frac{\delta_j}{3^j}+\frac{1}{3^N}<\sum_{j=1}^{N-1}\frac{\delta_j}{3^j}+\frac{2}{3^N}\le \sum_{j=1}^{\infty}\frac{\delta_j}{3^j}. \end{align*}\] This completes the proof.

Remarks

There is a bijection \(\phi:\{0,1\}^{\mathbb{N}} \to \mathcal{C}\) defined by \[\phi(z)=\frac{2}{3}\sum_{j=0}^{\infty}\frac{z_j}{3^j} \quad \text{ for } \quad z=(z_j)_{j \in \mathbb{N}}, \quad z_j \in \{0,1\},\] and consequently \({\rm card\;}(\mathcal{C})={\rm card\;}(\{0,1\}^{\mathbb{N}})={\rm card\;}(\mathbb{R})=\mathfrak{c}.\)

Cantor tree

Cantor–Lebesgue function

• Because \(C_n\) is a finite union of intervals, \[f_n(x)=\left(\frac{3}{2}\right)^n\lambda(C_n\cap[0, x]).\] is a polygonal function with \(f_n(0)=0\), \(f_n(1)=1\), and \(f_n\) is constant on each of \(2^{n}-1\) open intervals composing \([0,1] \setminus C_n\) and rises with slope \(\left(\frac{3}{2}\right)^n\) on each on the \(2^n\) closed intervals composing \(C_n\).

• If the \(j\)-th interval of \(C_n\) counting from the left is \([a_{n,j},b_{n,j}]\), then \[f_n(a_{n,j})=\frac{j-1}{2^n} \quad \text{ and } \quad f_n(b_{n,j})=\frac{j}{2^n}.\] Also \(a_{n,j}=a_{n+1, 2j-1}\) and \(b_{n, j}=b_{n+1, 2j}\) and \[f_{n+1}(a_{n, j})=f_{n}(a_{n, j}) \quad \text{ and } \quad f_{n+1}(b_{n, j})=f_{n}(b_{n, j})\] and \(f_{n+1}\) agrees with \(f_n\) at all of the endpoints of the intervals of \(C_n\) and therefore on \([0,1] \setminus C_n\).

• Within any particular interval \([a_{n,j},b_{n,j}]\) of \(C_n\) the greatest difference between \(f_n(x)\) and \(f_{n+1}(x)\) is at the new endpoints within that interval and the magnitude of the difference is at most \(\frac{1}{6 \cdot 2^n}\). Thus we have \[|f_{n+1}(x)-f_n(x)| \leq \frac{1}{6 \cdot 2^n} \quad \text{ for any } \quad n \in \mathbb{N}.\]

• Since \[\frac{1}{6}\sum_{n \in \mathbb{N}}\frac{1}{2^n}<\infty.\] thus \((f_n)_{n \in \mathbb{N}}\) is uniformly convergent to a continuous function \(f:[0,1] \to [0,1].\)

• This function is called the Cantor–Lebesgue function or Devil’s staircase function.

The Cantor–Lebesgue function or Devil’s staircase function:

Properties of the Cantor–Lebesgue function

• Every \(f_n\) is non-decreasing so is \(f\).

• Let \(\phi:\{0,1\}^{\mathbb{N}} \to \mathcal{C}\), be given by \[\phi(z)=\frac{2}{3}\sum_{j=0}^{\infty}\frac{z_j}{3^j}\] for every \(z=(z_j)_{j \in \mathbb{N}} \in \{0,1\}^{\mathbb{N}}\). Then \[f(\phi(z))=\frac{1}{2}\sum_{j=0}^{\infty}\frac{z_j}{2^j}.\]

• Indeed, fix \(\eta=\{\eta_0,\eta_1,\ldots\} \in \{0,1\}^{\mathbb{N}}\) and for each \(n \in \mathbb{N}\) take \(I_n\) to be the component of \(C_n\) containing \(\phi(\eta)\). Then \(I_{n+1}\) will be the left-hand third of \(I_{n+1}\) if \(\eta_n=0\) and the right-hand third if \(\eta_n=1\).

• Taking \(a_n\) to be the left-hand endpoint of \(I_n\) we see that \[a_{n+1}=a_{n}+\frac{2}{3}\cdot\frac{\eta_n}{3^{n}} \quad\text{ and }\quad f_{n+1}(a_{n+1})=f_n(a_n)+\frac{1}{2}\cdot \frac{\eta_n}{2^{n}}\] for each \(n\in\mathbb N\). Thus \[\phi(\eta)=\lim_{n \to \infty}a_n\] and \[f(\phi(z))=\lim_{n \to \infty}f_n(a_n)=\frac{1}{2}\sum_{j=0}^{\infty}\frac{\eta_j}{2^j}.\]

• In particular, \(f[\mathcal{C}]=[0,1]\) since any \(x\in[0, 1]\) is expressible as \[x=\sum_{j=0}^{\infty}\frac{\eta_j}{2^{j+1}}=f(\phi(\eta))\] for some \(z =(z_j)_{j \in \mathbb{N}} \in \{0,1\}^{\mathbb{N}}\).

The axiom of choice

Proof.

Take \(f \in \prod_{\alpha \in A}X_{\alpha} \neq \varnothing\). Define \[Y=f[A],\] then \[Y \cap X_{\alpha}=\{f(\alpha)\}\] since \(f(\alpha) \in X_{\alpha}\).

Some equivalence relation

Let \(\{\epsilon_{\alpha}: \alpha\in A\}\) be the set of all equivalence classes. The equivalence classes \(\epsilon_{\alpha}\) are either disjoint or coincide, and \([0,1]\) is the disjoint union of all equivalence classes, we write \[[0,1]=\bigcup_{\alpha \in A}\epsilon_\alpha,\] and \[\epsilon_\alpha \cap \epsilon_\beta=\varnothing \quad \text{ if } \quad \alpha \neq \beta.\]

Nonmeasurable subset of \(\mathbb R\)

Proof. Since \([0,1]=\bigcup_{\alpha \in A}\epsilon_\alpha\), by the axiom of choice there is a selector \[\mathcal{N}=\{x_{\alpha}: \alpha \in A\}\] for the family \((\epsilon_\alpha)_{\alpha \in A}\) such that \[{\rm card\;}(\mathcal{N} \cap \epsilon_{\alpha})=1 \quad \text{ for any }\quad \alpha \in A\] We claim that the set \(\mathcal{N}\) is non-measurable.

• Let \(\{r_k: k \in \mathbb{N}\}=\mathbb{Q} \cap [0,1]\). For each \(k \in \mathbb{N}\) consider \[\mathcal{N}_k=r_k+\mathcal{N}.\]

• We show that

  1. \(\mathcal{N}_k \cap \mathcal{N}_l = \varnothing\) if \(k \neq l\),

  2. \([0,1] \subseteq \bigcup_{k=1}^{\infty}\mathcal{N}_k \subseteq [-1,2]\).

• Proof of (i). If \(\mathcal{N}_k \cap \mathcal{N}_l \neq \varnothing\) for some \(l \neq k\), then there are \(\alpha,\beta\in A\) and \(r_k \neq r_l\) such that \[x_{\alpha}+r_k=x_{\beta}+r_l \iff x_{\alpha}-x_{\beta}=r_l-r_k.\] Consequently, \(\alpha \neq \beta\) and \(x_{\alpha} - x_{\beta}\in \mathbb{Q}\). Hence, \(x_{\alpha} \sim x_{\beta}\). But this contradicts the fact that \(\mathcal{N}\) contains only one representative of each equivalence class.

• Proof of (ii). Of course \(\bigcup_{k=1}^{\infty}\mathcal{N}_k \subseteq [-1,2]\). If \(x \in [0,1]\) then \(x \sim x_{\alpha}\) for some \(\alpha \in A\) and \(x-x_{\alpha}=r_k\) for some \(k \in \mathbb{N}\) thus \[x=x_\alpha+r_k \in r_k+\mathcal{N}=\mathcal{N}_k.\]

• If we assume that \(\mathcal{N}\) is measurable then \[\begin{align*} \lambda(\mathcal{N})=\lambda(r_k+\mathcal{N})=\lambda(\mathcal{N}_k) \quad \text{ for all } \quad k \in \mathbb{N}, \end{align*}\] by the translation invariance of Lebesgue measure. Now note that \[\begin{align*} 1=\lambda([0,1]) &\leq \lambda\bigg(\bigcup_{k=1}^{\infty}\mathcal{N}_k\bigg)\\ &=\sum_{k=1}^{\infty}\lambda(\mathcal{N}_k) \\ &\leq \lambda([-1,2])=3. \end{align*}\]

• This is a contradiction since neither \(\lambda(\mathcal{N})=0\) nor \(\lambda(\mathcal{N})=1\).

Kakeya sets

• In 1917 Kakeya asked for the smallest area of a region in the plane inside which a unit needle can be rotated by \(180^{\circ}\).

• Besicovitch proved in 1919 that there exist Kakeya sets in \(\mathbb R^2\) of Lebesgue measure zero, so no smallest area exists.

• Such sets are nevertheless large in the sense of Lecture 4: the Kakeya conjecture asserts that every Kakeya set in \(\mathbb R^d\) has Hausdorff dimension \(d\). This is classical for \(d=2\) and was recently proved for \(d=3\) by Wang and Zahl.

Bishop’s construction of a Kakeya set in \(\mathbb R^2\)

Proof:

• The construction below produces a compact set \(\mathcal K \subseteq \mathbb R^2\) with \(\lambda_2(\mathcal K)=0\) which contains a unit segment of every slope in \([0,1]\).

• The union of four copies of \(\mathcal K\) rotated by \(0, \frac{\pi}{4}, \frac{\pi}{2}\) and \(\frac{3\pi}{4}\) is then a compact Kakeya set of measure zero, since every direction of \(\mathbb R^2\) is obtained from a slope in \([0,1]\) by one of these four rotations.

Step 1. The sequence \((x_k)_{k=0}^{\infty}\). Let \[x_k=\begin{cases} 0 & \text{ if } k=0,\\ \frac{k}{2^n}-1 & \text{ if } k \in [2^n, 2^{n+1}) \text{ and } n \text{ is even},\\ 2-\frac{k}{2^n} & \text{ if } k \in [2^n, 2^{n+1}) \text{ and } n \text{ is odd}. \end{cases}\]

• Thus \[(x_k)_{k=0}^{\infty}=\Big(0, 0, 1, \frac{1}{2}, 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1, \frac{7}{8}, \frac{6}{8}, \frac{5}{8}, \frac{4}{8}, \ldots\Big)\] that is, the sequence traverses the dyadic rationals \((j2^{-n})_{j=0}^{2^n}\), reversing the order each time. In particular \((x_k)_{k=0}^{\infty}\) is dense in \([0,1]\).

• If \(k \in [2^n, 2^{n+1})\), then \[\varepsilon(k):=|x_{k+1}-x_k|=\frac{1}{2^n},\] so \(\varepsilon(k)\ge \varepsilon(k+1)\) for \(k \geq 1\) and \(\lim_{k\to\infty}\varepsilon(k)=0\).

• Moreover, the intervals \([x_k-\varepsilon(k), x_k+\varepsilon(k)]\) cover \([0,1]\) infinitely often: for every \(k \in \mathbb N\) and every \(b \in [0,1]\) there is \(N>k\) with \(b \in [x_N-\varepsilon(N), x_N+\varepsilon(N)]\).

Step 2. The function \(f\) and the set \(\mathcal K\). Let \(\{s\}=s-\lfloor s \rfloor\) denotes the fractional part of \(s\in\mathbb R\). For \(t \in [0,1]\) let \(f(t)=\lim_{k \to \infty}f_k(t)\), where \[f_k(t)=\sum_{j=1}^{k}\{2^{j}t\}\frac{x_{j-1}-x_j}{2^{j}}.\] Define the set \(\mathcal K=\{(\alpha, f(t)+\alpha t): \alpha, t \in [0,1]\}\).

• The series converges uniformly, since \(|\{2^{j}t\}(x_{j-1}-x_j)| \leq 1\) for every \(j \in \mathbb N\), so \(f\) is a well defined bounded function and consequently \(\mathcal K\) is a bounded subset of \(\mathbb R^2\).

• \(\mathcal K\) contains a unit segment of every slope \(t \in [0,1]\): for a fixed \(t \in [0,1]\) the points \((\alpha, f(t)+\alpha t)\) with \(\alpha \in [0,1]\) form the segment joining \((0, f(t))\) and \((1, f(t)+t)\), whose slope is \(t\) and whose length is \(\sqrt{1+t^2} \geq 1\).

• It remains to show that \(\lambda_2({\rm cl}(\mathcal K))=0\); the set \({\rm cl}(\mathcal K)\) is then compact, and it still contains a unit segment of every slope in \([0,1]\).

Step 3. On each dyadic interval the function \(f_k(t)+x_kt\) is constant.

• Fix \(k \in \mathbb N\) and let \(I\) be a component of \([0,1] \setminus 2^{-k}\mathbb Z\), that is, \(I=\big(\frac{m}{2^{k}}, \frac{m+1}{2^{k}}\big)\) for some \(m \in \{0, 1, \ldots, 2^{k}-1\}\).

• If \(t \in I\) and \(1 \leq j \leq k\), then \(\{2^{j}t\}=2^{j}t-\lfloor 2^{j}t \rfloor\) is affine on \(I\) with slope \(2^{j}\), hence \(f_k\) is affine on \(I\) and, by telescoping, \[f_k'(t)=\sum_{j=1}^{k}(x_{j-1}-x_j)=x_0-x_k=-x_k, \quad \text{ since } \quad x_0=0.\]

• Consequently \[f_k(t)+x_kt \quad \text{ is constant on each component of } \quad \ [0,1] \setminus 2^{-k}\mathbb Z.\]

• Note also that \[|f(t)-f_k(t)| \leq \sum_{i>k}\{2^{i}t\}\frac{|x_{i-1}-x_i|}{2^{i}} \leq \varepsilon(k)\sum_{i>k}\frac{1}{2^{i}}=\frac{\varepsilon(k)}{2^{k}},\] because \(|x_{i-1}-x_i|=\varepsilon(i-1) \leq \varepsilon(k)\) for every \(i>k\).

Step 4. The diameter estimate. Fix \(n \in \mathbb N\) and \(k \in [2^n, 2^{n+1})\), so that \(\varepsilon(k)=2^{-n}\), and let \(\alpha \in [0,1]\) satisfy \(|\alpha-x_k| \leq \varepsilon(k)\).

• For \(t\) in a component \(I\) of \([0,1] \setminus 2^{-k}\mathbb Z\) we write \[f(t)+\alpha t=\big(f(t)-f_k(t)\big)+\big(f_k(t)+x_kt\big)+t(\alpha-x_k).\]

• The first term varies by at most \(2\varepsilon(k)2^{-k}\) by Step 3, the second one is constant on \(I\) by Step 3, and the third one varies by at most \(|I|\,|\alpha-x_k| \leq 2^{-k}\varepsilon(k)\).

• Hence the image of \(I\) under \(t \mapsto f(t)+\alpha t\) has diameter at most \[2\varepsilon(k)2^{-k}+2^{-k}\varepsilon(k)=\frac{3}{2^{n+k}}.\]

• If moreover \(\alpha\) varies in an interval \(J \subseteq [x_k-\varepsilon(k), x_k+\varepsilon(k)]\) with \(|J|=2^{-n-k}\), then the third term changes by at most \(|t|\,|J| \leq 2^{-n-k}\), so \[\{f(t)+\alpha t: t \in I, \ \alpha \in J\} \quad \text{ is contained in a box of side length } \quad \frac{4}{2^{n+k}}.\]

Step 5. The covering argument. Let \(\mathcal K_J=\{(\alpha, f(t)+\alpha t): \alpha \in J, \ t \in [0,1]\}\) with \(n\), \(k\) and \(J\) be as in Step 4.

• There are \(2^{k}\) components \(I\) of \([0,1] \setminus 2^{-k}\mathbb Z\) and \(2^{k}+1\) remaining dyadic points, and each of them contributes to \(\mathcal K_J\) a set contained in a closed square of side \(4 \cdot 2^{-n-k}\), by Step 4. Hence \[\lambda_2^{*}(\mathcal K_J) \leq (2^{k+1}+1)\frac{16}{2^{2n+2k}}=O\Big(\frac{1}{2^{2n+k}}\Big).\]

• The interval \([x_k-\varepsilon(k), x_k+\varepsilon(k)]\) is the union of \(2^{k+1}\) such intervals \(J\), and as \(k\) runs over \([2^n, 2^{n+1})\) the points \(x_k\) run over all the dyadic rationals \(j2^{-n}\), so these \(2^{n}\) intervals cover \([0,1]\). Therefore \[\lambda_2^{*}(\mathcal K) \leq 2^{n} \cdot 2^{k+1} \cdot O\Big(\frac{1}{2^{2n+k}}\Big) =O\Big(\frac{1}{2^{n}}\Big).\]

• For every \(n\) the set \(\mathcal K\) is covered by finitely many closed squares of total area \(O(2^{-n})\), so the same is true for \({\rm cl}(\mathcal K)\), and \(\lambda_2({\rm cl}(\mathcal K))=0\) by letting \(n \to \infty\).

• Finally, \({\rm cl}(\mathcal K)\) is compact and the union of its four copies rotated by \(0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}\) is a compact Kakeya set of Lebesgue measure zero.

The set \(\mathcal K\)

• The segments joining \((0, f(t))\) and \((1, f(t)+t)\) for \(160\) values of \(t \in [0,1]\); the whole set \(\mathcal K\) is the union of all of them, and it has Lebesgue measure zero.

Four rotated copies of \(\mathcal K\)

• Rotating \(\mathcal K\) by \(0, \frac{\pi}{4}, \frac{\pi}{2}\) and \(\frac{3\pi}{4}\) we obtain unit segments in every direction, while the union still has Lebesgue measure zero.

Generalized Cantor sets

• For \(\alpha_j=\frac{1}{3}\) we recover the ternary Cantor set from this lecture.

Lebesgue measurable sets which are not Borel

Density of measurable sets and Steinhaus’ theorem

Perfect subsets of sets of positive measure

Two consequences

• Problem 9 also follows immediately from Steinhaus’ theorem.