# Totally disconnected compact set with positive measure

Let’s build a totally disconnected compact set $$K \subset [0,1]$$ such that $$\mu(K) >0$$ where $$\mu$$ denotes the Lebesgue measure.

In order to do so, let $$r_1, r_2, \dots$$ be an enumeration of the rationals. To each rational $$r_i$$ associate the open interval $$U_i = (r_i – 2^{-i-2}, r_i + 2^{-i-2})$$. Then take $\displaystyle V = \bigcup_{i=1}^\infty U_i \text{ and } K = [0,1] \cap V^c.$ Clearly $$K$$ is bounded and closed, therefore compact. As Lebesgue measure is subadditive we have $\mu(V) \le \sum_{i=1}^\infty \mu(U_i) \le \sum_{i=1}^\infty 2^{-i-1} = 1/2.$ This implies $\mu(K) = \mu([0,1]) – \mu([0,1] \cap V) \ge 1/2.$ In a further article, we’ll build a totally disconnected compact set $$K^\prime$$ of $$[0,1]$$ with a predefined measure $$m \in [0,1)$$.

# A strictly increasing continuous function that is differentiable at no point of a null set

We build in this article a strictly increasing continuous function $$f$$ that is differentiable at no point of a null set $$E$$. The null set $$E$$ can be chosen arbitrarily. In particular it can have the cardinality of the continuum like the Cantor null set.

### A set of strictly increasing continuous functions

For $$p \lt q$$ two real numbers, consider the function $f_{p,q}(x)=(q-p) \left[\frac{\pi}{2} + \arctan{\left(\frac{2x-p-q}{q-p}\right)}\right]$ $$f_{p,q}$$ is positive and its derivative is $f_{p,q}^\prime(x) = \frac{2}{1+\left(\frac{2x-p-q}{q-p}\right)^2}$ which is always strictly positive. Hence $$f_{p,q}$$ is strictly increasing. We also have $\lim\limits_{x \to -\infty} f_{p,q}(x) = 0 \text{ and } \lim\limits_{x \to \infty} f_{p,q}(x) = \pi(q-p).$ One can notice that for $$x \in (p,q)$$, $$f_{p,q}^\prime(x) \gt 1$$. Therefore for $$x, y \in (p,q)$$ distinct we have according to the mean value theorem $$\frac{f_{p,q}(y)-f_{p,q}(x)}{y-x} \ge 1$$.

### Covering $$E$$ with an appropriate set of open intervals

As $$E$$ is a null set, for each $$n \in \mathbb N$$ one can find an open set $$O_n$$ containing $$E$$ and measuring less than $$2^{-n}$$. $$O_n$$ can be written as a countable union of disjoint open intervals as any open subset of the reals. Then $$I=\bigcup_{m \in \mathbb N} O_m$$ is also a countable union of open intervals $$I_n$$ with $$n \in \mathbb N$$. The sum of the lengths of the $$I_n$$ is less than $$1$$. Continue reading A strictly increasing continuous function that is differentiable at no point of a null set

# Small open sets containing the rationals

The set $$\mathbb Q$$ of the rational number is countable infinite and dense in $$\mathbb R$$. You can have a look here on a way to build a bijective map between $$\mathbb N$$ and $$\mathbb Q$$.

Now given $$\epsilon > 0$$, can one find an open set $$O_\epsilon$$ of measure less than $$\epsilon$$ with $$\mathbb Q \subseteq O_\epsilon$$?

The answer is positive. Let’s denote $O_\epsilon = \bigcup_{n \in \mathbb N} (r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}})$ where $$(r_n)_{n \in \mathbb N}$$ is an enumeration of the rationals. Obviously $$\mathbb Q \subseteq O_\epsilon$$. Using countable subadditivity of Lebesgue measure $$\mu$$, we get:
\begin{align*}
\mu(O_\epsilon) &\le \sum_{n \in \mathbb N} \mu((r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}}))\\
&= \sum_{n \in \mathbb N} \frac{2 \epsilon}{2^{n+1}} = \sum_{n \in \mathbb N} \frac{\epsilon}{2^n} = \epsilon
\end{align*}

Therefore we’re done. Some additional comments:

• While Lebesgue measure of the reals is infinite and the rationals are dense in the reals, we can include the rationals in an open set of measure as small as desired!
• The open segments $$(r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}})$$ are overlapping. Hence $$\mu(O_\epsilon)$$ is strictly less than $$\epsilon$$.

# A non-measurable set

We describe here a non-measurable subset of the segment $$I=[0,1] \subset \mathbb R$$.

Let’s define on $$I$$ an equivalence relation by $$x \sim y$$ if and only if $$x-y \in \mathbb Q$$. The equivalence relation $$\sim$$ induces equivalence classes on $$I$$. For $$x \in I$$, it’s equivalence class $$[x]$$ is $$[x] = \{y \in I \ : \ y-x \in \mathbb Q\}$$. By the Axiom of Choice, we can form a set $$A$$ by selecting a single point from each equivalence class.

We claim that the set $$A$$ is not Lebesgue measurable.

For all $$q \in \mathbb Q$$ we denote $$A_q = \{q+x \ : x \in A\}$$. Let’s take $$p,q \in \mathbb Q$$. If it exists $$z \in A_p \cap A_q$$, it means that there exist $$u,v \in A$$ such that
$z= p+u=q+v$ hence $$u-v=q-p=0$$ as $$u,v$$ are supposed to be unique representatives of the classes of the equivalence relation $$\sim$$. Finally if $$p,q$$ are distincts, $$A_p \cap A_q = \emptyset$$.

As Lebesgue measure $$\mu$$ is translation invariant, we have for $$q \in \mathbb Q \cap [0,1]$$ : $$\mu(A) = \mu(A_q)$$ and also $$A_q \subset [0,2]$$. Hence if we denote
$B = \bigcup_{q \in \mathbb Q \cap [0,1]} A_q$ we have $$B \subset [0,2]$$. If we suppose that $$A$$ is measurable, we get
$\mu(B) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A) \le 2$ by countable additivity of Lebesgue measure (the set $$\mathbb Q \cap [0,1]$$ being countable infinite). This implies $$\mu(A) = 0$$.

Let’s prove now that
$[0,1] \subset \bigcup_{q \in \mathbb Q \cap [-1,1]} A_q$ For $$z \in [0,1]$$, there exists $$u \in A$$ such that $$z \in [u]$$. As $$A \subset [0,1]$$, we have $$q = z-u \in \mathbb Q$$ and $$-1 \le q \le 1$$. And $$z=q+u$$ means that $$z \in A_q$$. This proves the inclusion. However the inclusion implies the contradiction
$1 = \mu([0,1]) \le \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A) =0$

Finally $$A$$ is not Lebesgue measurable.

# The Smith Volterra Cantor Set

In Cantor set article, I presented the Cantor set which is a null set having the cardinality of the continuum. I present here a modification of the Cantor set named the Smith-Volterra-Cantor set.

### Construction of the Smith-Volterra-Cantor set

The Smith-Volterra-Cantor set (also named SVC set below) $$S$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$S_0=I$$
• $$S_1=[0,\frac{3}{8}] \cup [\frac{5}{8},1]$$
• If $$S_n$$ is a finite disjoint union of segments $$s_n=\cup_k \left[a_k,b_k\right]$$, $S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)$

# Counterexamples around Fubini’s theorem

We present here some counterexamples around the Fubini theorem.

We recall Fubini’s theorem for integrable functions:
let $$X$$ and $$Y$$ be $$\sigma$$-finite measure spaces and suppose that $$X \times Y$$ is given the product measure. Let $$f$$ be a measurable function for the product measure. Then if $$f$$ is $$X \times Y$$ integrable, which means that $$\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) < \infty$$, we have $\int_X \left( \int_Y f(x,y) dy \right) dx = \int_Y \left( \int_X f(x,y) dx \right) dy = \int_{X \times Y} f(x,y) d(x,y)$ Let's see what happens when some hypothesis of Fubini's theorem are not fulfilled. Continue reading Counterexamples around Fubini’s theorem

# Cantor set: a null set having the cardinality of the continuum

## Definition of the Cantor set

The Cantor ternary set (named Cantor set below) $$K$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$K_0=I$$
• If $$K_n$$ is a finite disjoint union of segments $$K_n=\cup_k \left[a_k,b_k\right]$$, $K_{n+1}=\bigcup_k \left(\left[a_k,a_k+\frac{b_k-a_k}{3}\right] \cup \left[a_k+2\frac{b_k-a_k}{3},b_k\right]\right)$

And finally $$K=\displaystyle \bigcap_{n \in \mathbb{N}} K_n$$. The Cantor set is created by repeatedly deleting the open middle third of a set of line segments starting with the segment $$I$$.

The Cantor set is a closed set as it is an intersection of closed sets. Continue reading Cantor set: a null set having the cardinality of the continuum