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## 2019 Fall: Real analysis I (M721)

### Some interesting results

There exits sets which are not Lebesgue measurable. One standard construction is based on the axiom of choice. However, Solovay showed that the axiom of choice is essential to the proof of the existence of a non-measurable set. He constructed a model in which all of the axioms of Zermelo–Fraenkel set theory hold, exclusive of the axiom of choice, but in which all sets of real numbers are Lebesgue measurable. The construction relies on the existence of an inaccessible cardinal. Cited from the wikipedia page of Solovay model.

The statement, Each linear operator on a Hilbert space is a bounded linear operator, is consistent with ZF without the axiom of choice. Indeed, the above statement can not be disproved either in ZF+some weekended form of the axiom of choice.

This was discovered by J. Wright in "All operators on a Hilbert space are bounded", Bull. Amer. Math. Soc. 79 (1973), 1247-1250. The argument is built on Solovay's work mentioned above.

I learnt this paper from Betsy Stovall.

The cardinality of the collection of Borel sets on $\mathbb{R}$ is $2^{\aleph_0}$. The cardinality of the collection of Lebesgue sets is $2^{2^{\aleph_0}}.$ Cantor sets already have the cardinality of $2^{\aleph_0}$. Moreover, every subset of a Cantor set is Lebesgue measurable.

Let $f$ be a real-valued Lebesgue measurable function defined on the interval $[0, 1]$. Then there exists a sequence of positive real numbers $\{h_k\}$ with $h_k\to 0$ such that
$$f(x+h_k)\to f(x), \text{ almost everywhere.}$$

The standard Cantor function $C: [0, 1]\to [0, 1]$ is often useful in constructing counterexamples. Based on it, one can also construct a strictly increasing continuous function whose derivative vanishes almost everywhere. The following construction is given in Page 109 of the book by Folland. Let $\{[a_n, b_n]\}$ be an enumeration of intervals inside $[0, 1]$ with rational endpoints. Given $C$ a trivial extension to the domain $\mathbb{R}$ and then consider
$$G=\sum_{n=1}^{\infty} C\left(\frac{x-a_n}{b_n-a_n}\right).$$
By a similar construction as above, one can construct a function from $\mathbb{R}$ to itself that is discontinuous at and only at rational points.

In the definition of the Hausdorff measure: Let $p\ge 0$ and $\delta>0$. For $A\subset \mathbb{R}^n$, let
$$H_{p, \delta}(A)=\inf\{\sum_{j=1}^{\infty}(\text{diam} B_j)^p: A\subset \bigcup_{j=1}^{\infty} B_j \text{ and } \text{ diam} B_j\le \delta\}.$$
Then
$$H_p(A):=\lim_{\delta\to 0} H_{p, \delta}(A)$$
Notice that here $B_j$ can be an arbitrary set. However, if we restrict $B_j$ to be a ball, then we may obtain a different measure (though still comparable). See Page 681 of the book "Real analysis: theory of measure and integration" by Yeh. I did not find any example there.

If $E$ is a subset of $\mathbb{R}^n$ we define the distance set
$$D(E)=\{|x-y|: x, y\in E\}.$$
If $E$ is a measurable set with positive Lebesgue measure, then $D(E)$ contains an interval around the origin. This is a result due to Steinhaus and we had it as an exercise.

In $\mathbb{R}^n$ for every $n\ge 1$, even if $\text{dim}(E)=n$, the distance set $D(E)$ may not contain any interval with left end-point zero. However it is not known whether $D(E)$ contains some other non-degenerate intervals if $\text{dim}(E)$ is sufficiently close to $n$. (See Mattila's book Page 165.)

Theorem [Falconer 85]: On the plane, for every $s\in (0, 1)$, there exists a compact set $E\subset \mathbb{R}^2$ with $\text{dim}(E)=s$ and $\text{dim}(D(E))\le s.$

Let $E\subset \mathbb{R}$ be a measurable set. We say that $E$ is mid-point convex if for every $x, y\in E$, we also have $\frac{x+y}{2}\in E$. Prove that if $E$ is mid-point convex and $|E|>0$, then $E$ is convex.

First of all, by assumption, we have $\frac{E+E}{2}\subset E$. Secondly, by applying the Lebesgue differentiation theorem， it is not difficult to see that $E+E$ contains an open interval. Hence $E$ itself contains an open interval. This further implies that $E$ is convex.

However, there exists a "large" set that is mid-point convex but not convex. Let $\{v_{\alpha}\}$ be a $\mathbb{Q}$-basis for $\mathbb{R}$, then
$$A=\{q_1 v_{\alpha_1}+q_2 v_{\alpha_2}+\dots +q_n v_{\alpha_n}: n\ge 1, q_i\in \mathbb{Q}, \sum q_i\ge 0\}$$
and
$$B=\{q_1 v_{\alpha_1}+q_2 v_{\alpha_2}+\dots +q_n v_{\alpha_n}: n\ge 1, q_i\in \mathbb{Q}, \sum q_i< 0\}$$
form a partition of $\mathbb{R}$. Both $A$ and $B$ are mid-point convex. It is not difficult to see that neither $A$ nor $B$ is convex. (I learnt this from a solution Thomas Kragh gave at mathoverflow.)

However, the answer here may not be satisfying because neither $A$ nor $B$ is measurable. (Otherwise, by the previous argument, we would be able to conclude that both $A$ and $B$ are convex. )