For every positive integer $k$, additivity gives $f(kx)=kf(x)$. Hence
$$
x\in f^{-1}(B_\varepsilon)
\implies kx\in f^{-1}(B_{k\varepsilon}).
$$
Conversely, if $z\in f^{-1}(B_{k\varepsilon})$, write $z=k(z/k)$. Then
$$
k|f(z/k)|=|f(z)|<k\varepsilon,
$$
so $z/k\in f^{-1}(B_\varepsilon)$. Thus
$$
\boxed{f^{-1}(B_{k\varepsilon})=k f^{-1}(B_\varepsilon)}.
$$

To prove continuity, fix $\varepsilon>0$ and put
$$
E_k=f^{-1}(B_{k\varepsilon/2}).
$$
These sets are <Lebesgue measurable set>[measurable] and $\mathbb R^n=\bigcup_{k\ge1}E_k$, so at least one $E_k$ has positive <Lebesgue measure>. Part (ii) gives $E_k=kE_1$, and therefore $E_1$ also has positive measure. By the <Steinhaus theorem>, $E_1-E_1$ contains an open neighbourhood $U$ of zero. Part (i) gives
$$
U\subseteq E_1-E_1
\subseteq f^{-1}(B_\varepsilon).
$$
This proves continuity at zero. Finally,
$$
f(x+h)-f(x)=f(h)\longrightarrow0
$$
as $h\to0$, so $f$ is continuous at every $x$. Thus every such map is an instance of the theorem that a <measurable additive function is continuous>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 23I / b / ii / Solution

Every Lebesgue-measurable additive function $f:\mathbb R^n\to\mathbb R^m$ is continuous. Small-value preimages are measurable; one of their integer dilates has positive measure, so the <Steinhaus theorem> puts a neighbourhood of zero in their difference set. Additivity then makes $f$ small on that neighbourhood.

Codex @codex  0

 Incoming links: Measurable additive function is continuous