Exercise 1.6.9 (Measurable group homomorphisms are continuous)

This is a difficult exercise. Before we start, we will need a small lemma.

We now proceed with the proof.

(i)

We build the proof step-by-step, but maybe it is easier to read it in the reverse order.

• There exists a $z\in ℂ$ such that $f^{-1}(z+B(0,r))$ has positive measure.
  We exclude the degenerate case where everything maps to zero, and all of the above conditions are satisfied anyway. Thus, assume that there exists a $x\in {ℝ}^d$ such that $f(x)\ne 0$.
• Consider $E:=f^{-1}(z+B(0,r))$. Then the set $E-E$ contains a ball centred at origin $B_{{ℝ}^d}(0,\delta )$. Notice that the image of this ball is bounded, since $x\in B⟹x=e_1-e_2$ for some $e_1,e_2\in E$ $⟹f(x)=f(e_1-e_2)=f(e_1)-f(e_2)=(z+y_1)-(z+y_2)=y_1-y_2$ for some $y_1,y_2\in B_{ℂ}(0,r)$. In other words, we have shown that $x\in B⟹f(x)\in B_{ℂ}(0,2r)$, i.e., $f(B_{{ℝ}^d}(0,\delta ))\subseteq B_{ℂ}(0,2r)$. We make use of this fact.
• If the image of $B(0,\delta )$ is bounded and $f$ scales linearly, then both $B(0,\frac{1}{k}\delta )$ and the image of $B(0,\frac{1}{k}\delta )$ under $f$ must shrink as $k\to \infty$.
  Notice that the set $B_{{ℝ}^d}(0,\frac{1}{k}\delta )$ becomes increasingly small as $k\to \infty$. Pick an arbitrary ${(x_n)}_{n\in \mathbb{N}}$ in ${ℝ}^d$ which converges to 0. Then we can find a $k\in \mathbb{N}$ big enough so that $\parallel x\parallel \le k⟹x\in B_{{ℝ}^d}(0,\frac{1}{k}\delta )$. But $f(B(0,\delta ))$ is bounded, and so by $f(\frac{1}{n}x)=\frac{1}{n}f(x)$ we know that $B_{{ℝ}^d}(0,\frac{1}{k}\delta )\subseteq B_{ℂ}(0,\frac{1}{k}2r)$. Thus, we see that $f{(x_n)}_{n\in \mathbb{N}}$ is bound to converge to zero, too. Thus $f$ is continuous at the origin.
• If $f$ is continuous at the origin, then $f$ is continuous everywhere.
  Let ${(x_n)}_{n\in \mathbb{N}}$ be a sequence of points in ${ℝ}^d$ such that ${(x_n)}_{n\in \mathbb{N}}\to x_0$. We have to demonstrate that $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x_n)=f(x_0)$. In other words, $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x_n)-f(x_0)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x_n-x_0)=0$. But ${(x_n-x_0)}_{n\in \mathbb{N}}$ is a sequence converging to zero, and we are done.

• BONUS: $\forall <!--\; FUNCTION\; APPLICATION\; -->c\in ℝ:f(\mathit{cx})=\mathit{cf}(x)$.
  Let ${(q_n)}_{n\in \mathbb{N}}$ be a sequence of rational numbers converging to $c$. By continuity of $f$ we then have

  $$f(\mathit{cx})=f(\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{q}_{n}x)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(q_{n}x)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{q}_{n}f(x)=\mathit{cf}(x).$$

(ii)

Let $x\in {\mathbb{Q}}^d$. We then have $$f\left(\left[\begin{array}{c}\hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_d\hfill \end{array}\right]\right)=f\left(q_1\left[\begin{array}{c}\hfill 1\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]+\dots +q_d\left[\begin{array}{c}\hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 1\hfill \end{array}\right]\right)=q_{1}f\left(\left[\begin{array}{c}\hfill 1\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]\right)+\dots +q_{d}f\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 1\hfill \end{array}\right]\right).$$

We set $z_1:=f(e_1),\dots ,z_d=f(e_d)$, and the result follows for rational $q\in {\mathbb{Q}}^d$. Since each $x\in {ℝ}^d$ is a limit of some sequence ${(q_n)}_{n\in \mathbb{N}}$ of points in ${\mathbb{Q}}^d$, the overall result follows by taking the limits.