9 Cauchy-Hamel functional equation

9.1 Positive measure additive subgroups of the reals

Lemma 9.1 Countably many connected components for an open set

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Let $X$ be a locally connected separable space. Then any open subset $U \subseteq X$ has at most countably many connected components.

Proof ▶

(The proof is already formalized, see: IsOpen.countable-setOf-connectedComponentIn.)

For subsets $A, B \subseteq \mathbb {R}$, we use the following notation for pointwise sum sets and difference sets:

\begin{align*} A + B & = \left\{ a + b \; \big| \; a \in A, \, b \in B \right\} , \\ A - B & = \left\{ a - b \; \big| \; a \in A, \, b \in B \right\} . \end{align*}

We denote by $\Lambda $ the Lebesgue measure on $\mathbb {R}$.

Lemma 9.2 Finding an interval with high overlap

Let $A \subset \mathbb {R}$ be a measurable set such that $0 {\lt} \Lambda [A] {\lt} +\infty $. Then for any $r \in [0,1)$, there exists a nontrivial interval $J \subset \mathbb {R}$ (a subset of the real line which is connected and has nonempty interior) such that

\begin{align*} \Lambda [A \cap J] \; {\gt} \; r \, \Lambda [J] . \end{align*}

Proof ▶

Assume, without loss of generality, $0 {\lt} r {\lt} 1$. Since the Lebesgue measure is outer regular, we can find an open set $U \subset \mathbb {R}$ such that $A \subseteq U$ and $\Lambda [U] {\lt} r^{-1} \, \Lambda [A]$.

The open set $U$ has at most countably many connected components (which are in fact open intervals); denote by $(U_i)_{i \in I}$ the indexed collection of them.

Note that for at least one index $j \in I$ we have $\Lambda [A \cap U_j] {\gt} r \Lambda [U_j]$, because otherwise we get

\begin{align*} \Lambda [A] = \Lambda [A \cap U] = \, & \sum _{i \in I} \Lambda [A \cap U_i] \\ \le \, & r \sum _{i \in I} \Lambda [U_i] \\ = \, & r \Lambda [U], \end{align*}

contradicting the choice of $U$.

Now $J = U_j$ has the desired properties.

Lemma 9.3 Shifts of a smaller interval contained in a larger interval

Let $I, J$ be nontrivial intervals, whose length satisfy

\begin{align*} 0 \; {\lt} \; \Lambda [J] \; {\lt} \; \Lambda [I] \; {\lt} \, +\infty . \end{align*}

Then there exists an open interval $\Delta $ of length $\Lambda [\Delta ] = \Lambda [I] - \Lambda [J] {\gt} 0$ such that

\begin{align*} t + J \; \subset \; I \qquad \text{ for any $t \in \Delta $.} \end{align*}

Proof ▶

…

Lemma 9.4 Overlapping union of copies of an interval

Let $J$ be a nontrivial interval of finite length ($0 \, {\lt} \, \Lambda [J] \, {\lt} +\infty $). Let $c {\lt} 1$ and denote $\delta = c \Lambda [J] \, {\lt} \, \Lambda [J]$. Then for any $t \in (-\delta ,\delta )$, the set $J' = (t+J) \cup J$ is an interval (connected set with nonempty interior) whose length satisfies the bound $\Lambda [J'] \, {\lt} \, (1+c) \, \Lambda [J]$.

Proof ▶

Denote $a = \inf J$ and $b = \sup J$. We have $-\infty {\lt} a {\lt} b {\lt} +\infty $ and

\begin{align*} (a,b) \; \subseteq \; J \; \subseteq \; [a,b] \end{align*}

and we have $\Lambda [J] = b - a$.

Let $t \in (+\delta ,\delta )$, with $\delta = c \, \Lambda [J] = c(b-a)$. Then we have the containments

\begin{align*} \big( \min \{ a,a+t\} , \max \{ b,b+t\} \big) \; \subseteq \; (t+J) \cup J \; \subseteq \; \big[ \min \{ a,a+t\} , \max \{ b,b+t\} \big] . \end{align*}

It in particular follows that $J' := (t+J) \cup J$ is an interval.

If $t \ge 0$, then $J'$ is contained in $[a,b+t]$ which has length $b+t-a = (b-a) + t {\lt} \Lambda [J] + c \, \Lambda [J] = (1+c) \Lambda [J]$. If $t {\lt} 0$, then $J'$ is contained in $[a+t,b]$ and one similarly gets a length bound.

Lemma 9.5 Difference set of positive measure set contains an interval

Let $A \subset \mathbb {R}$ be a measurable set of positive Lebesgue measure. Then there exists a $\delta {\gt} 0$ such that

\begin{align*} (-\delta ,\delta ) \, \subseteq \, A - A . \end{align*}

Proof ▶

Pick a measurable subset $A_0 \subseteq A$ such that $0 {\lt} \Lambda [A_0] {\lt} +\infty $.

By Lemma 9.2, we can find a nontrivial interval $J$ such that $\Lambda [A_0 \cap J] {\gt} \frac{3}{4} \Lambda [J]$. Let $\delta = \frac{1}{2} \Lambda [J]$. We claim that $(-\delta ,\delta ) \subseteq A_0 - A_0$ (which then clearly implies the assertion of the lemma). Indeed, suppose $t \in (-\delta ,\delta )$. Then by Lemma 9.4, $(t+J) \cup J$ is an interval of length less than $\frac{3}{2} \Lambda [J]$. Moreover, we have $A_0 \cap J \subseteq (t+J) \cup J$ and $t+(A_0 \cap J) \subseteq (t+J) \cup J$. Now note that

\begin{align*} \Lambda [t + (A_0 \cap J)] = \Lambda [A_0 \cap J] {\gt} \frac{3}{4} \Lambda [J] . \end{align*}

If the sets $t + (A_0 \cap J)$ and $(A_0 \cap J)$ were disjoint, then the measure of $(A_0 \cap J) \cup (t+(A_0 \cap J))$ would thus be greater than $\frac{3}{2} \Lambda [J]$, which is impossible given the length of the interval $(t+J) \cup J$ is less than $\frac{3}{2} \Lambda [J]$. Therefore there exists a point $a \in (t + (A_0 \cap J)) \cap (A_0 \cap J)$. Denoting $a' = a - t$, we have $a, a' \in A_0 \cap J \subseteq A_0$ and $t = a - a' \in A_0 - A_0 \subseteq A - A$. Since $t \in (-\delta ,\delta )$ was arbitrary, this proves the assertion.

Lemma 9.6 Dividing a high overlap interval

Let $A$ be a measurable. Suppose that for some $a {\lt} b$, the interval $J = (a,b)$ satisfies $\Lambda [A \cap J] \; {\gt} \; r \, \Lambda [J]$. Let $m \in \mathbb {N}_{+}$, and consider the subintervals

\begin{align*} J_i = \Big( a + \frac{i}{(b-a)m} , \; a + \frac{i+1}{(b-a)m} \Big) \qquad \text{ for $i = 0, \ldots , m-1$.} \end{align*}

Then for some $i$ we have $\Lambda [A \cap J_i] \; {\gt} \; r \, \Lambda [J_i]$.

Proof ▶

Lemma 9.7 Difference of two positive measure sets contains an interval

Let $A, B \subset \mathbb {R}$ be two measurable sets of positive Lebesgue measure. Then the difference set $A - B$ contains a nontrivial open interval.

Proof ▶

Pick measurable subsets $A_0 \subseteq A$ and $B_0 \subseteq B$ such that $0 {\lt} \Lambda [A_0] {\lt} +\infty $ and $0 {\lt} \Lambda [B_0] {\lt} +\infty $.

By Lemma 9.2, we can find nontrivial intervals $J, I$ such that $\Lambda [A_0 \cap J] {\gt} \frac{3}{4} \Lambda [J]$ and $\Lambda [B_0 \cap I] {\gt} \frac{3}{4} \Lambda [I]$.

By Lemma 9.6 for any $n,m \in \mathbb {N}$ we can find intervals $I' \subset I$ and $J' \subset J$ with $\Lambda [I'] = \frac{1}{n} \Lambda [I]$ and $\Lambda [J'] = \frac{1}{m} \Lambda [J]$ and such that $\Lambda [A_0 \cap J'] {\gt} \frac{3}{4} \Lambda [J']$ and $\Lambda [B_0 \cap I'] {\gt} \frac{3}{4} \Lambda [I']$. By choosing $n$ and $m$ suitably, we can ensure that

\begin{align*} \frac{1}{2} \Lambda [I’] \; \le \; \Lambda [J’] \; {\lt} \; \Lambda [I’] . \end{align*}

Then by Lemma 9.3 there exists an open interval $\Delta $ of length $\Lambda [\Delta ] \, {\gt} \, \Lambda [I'] - \Lambda [J']$ such that for all $t \in \Delta $ we have $t + J' \subset I'$.

Consider a fixed $t \in \Delta $. Observe that $t + (B_0 \cap J) \subset I'$ and

\begin{align*} \Lambda [t + (B_0 \cap J’)] \; = \; \Lambda [B_0 \cap J’] \; {\gt} \; \frac{3}{4} \Lambda [J’] \; \ge \; \frac{3}{8} \Lambda [I’] . \end{align*}

We now claim that $A_0 \cap I'$ and $t + (B_0 \cap J')$ intersect. Their measures add up to at least

\begin{align*} \Lambda [A_0 \cap I’] + \Lambda [t + (B_0 \cap J’)] \; = \; & \Lambda [A_0 \cap I’] + \Lambda [B_0 \cap J’] \\ \; {\gt} \; & \frac{3}{4} \Lambda [I’] + \frac{3}{4} \Lambda [J’] \\ \; \ge \; & \frac{3}{4} \Lambda [I’] + \frac{3}{8} \Lambda [I’] \; = \; & \frac{9}{8} \Lambda [I’] . \end{align*}

and both have been shown to be subsets of $I'$; therefore they cannot be disjoint. In particular there is a point $z \in (A_0 \cap I') \cap (t + B_0 \cap J')$, which means that

\begin{align*} a = z = t + b \end{align*}

for some $a \in A_0 \cap I'$ and $b \in B_0 \cap J'$. We solve $t = a - b \in A_0 - B_0 \subseteq A - B$. Since $t \in \Delta $ was arbitrary, we have shown that the nontrivial open interval $\Delta $ is contained in $A - B$.