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I am trying to solve the following exercise:

Let $f:\mathbb{R}^d\to\mathbb{C}$ be a group homomorphism between $(\mathbb{R}^d,+)$ and $(\mathbb{C},+)$, i.e., $f(x+y) = f(x)+f(y)$. Show that if $f$ is Lebesgue measurable, then $f$ is continuous.

The hint to the exercise says to demonstrate that for any open ball $B_{\mathbb{C}}(0,r)$ centred at the origin of the complex plane, there is at least one $z\in\mathbb{C}$ such that the set $f^{-1}(z+B_{\mathbb{C}}(0,r))$ has positive measure. One then has to use Steinhaus theorem, which states that for any $E\subseteq \mathbb{R}^d: \mu(E)>0$ the set $E-E$ contains a neighbourhood of the origin.

Unfortunately, I do not understand how the hint relates to the exercise (i.e., how this leads to "an inverse image of any open ball/set is an open set). What I have done so far is:

  1. If $f^{-1}(z+B(0,r))$ is null for every $z$, then all of $\mathbb{R}^d$ is null by expressing the latter type as a countable union of sets of former type. Thus, there is at least one such $z$.
  2. $f(0) = 0$, so we use the origin as a starting point.
qarabala
  • 998

2 Answers2

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It is straightforward to check that $f(rx) = r f(x)$ for any rational $r$.

Suppose that in some neighbourhood $B(0, \delta)$, the function $f$ is bounded by $K$. Then we see that if $x \in B(0, {1 \over n} \delta)$ we have $f(x) \in B(0, {1 \over n} K)$. In particular, $f$ is continuous at the origin.

Since $f(x+y) = f(x)+f(y)$, we see that if $f$ is continuous at the origin, it is continuous at any $y \in \mathbb{R}^n$.

As an aside, if $r_n \to \lambda$ where $r_n$ are rationals, we see that $f(\lambda x) = \lim_n f(r_nx) = \lim_n r_n f(x) = \lambda f(x)$, hence $f$ is linear.

To see where the neighbourhood of the origin comes from, suppose $E= f^{-1}(z+B(0,r))$ has positive measure. Then $E-E$ contains some neighbourhood of the origin, say $B(0,\delta)$.

Hence if $x \in B(0,\delta) \subset E-E$, we have $x = e_1 -e_2$ for some $e_k \in E$. Then $f(e_k) = z+z_k$ where $z_k \in B(0,r)$, and so $f(x) = z_1-z_2 \in B(0,2r)$.

copper.hat
  • 172,524
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Proof:

Fix any $r > 0$. Let $D = B_{\mathbb{C}}(0,r)$ and $B \subset \mathbb{C}$ be any rational box of $\mathbb{C}$. Then we can write the complex plane as a countable cover

$\mathbb{C} = \bigcup_{B; B \subset z + D \text{ for some }z\in \mathbb{C}} z + D$

by the shifted disk $D$.(This is because any open disk $B_{\mathbb{C}}(x,r)$ in $\mathbb{C}$ contains a rational box $B$ such that $x \in B$, see for example Lemma $7.1.10$ of Analysis $2$ by Terence Tao). Hence by contradiction if $f^{-1}(z + D)$ is a null set for any $z \in \mathbb{C}$, we would have $\mathbb{R}^d = f^{-1}(\mathbb{C}) = \bigcup_{B; B \subset z + D \text{ for some }z\in \mathbb{C}} f^{-1}(z + D)$ is a null set, a contradiction. So $f^{-1}(z + D)$ has positive measure for at least one $z \in \mathbb{C}$.

Fix this $z$ and denote $E = z + D$. By Steinhaus theorem we see that $E - E$ contains an open neighborhood $U$ of zero. Hence $\forall x_0 \in \mathbb{R}^d$, the quantity $x - x_0$ will be such that $x - x_0 \in U$ for $|x - x_0|$ sufficiently small. Then $x - x_0 = r - s$ for some $r, s \in E$. By construction, $f(r) = z + r'$ for some $r' \in D$, and similarly $f(s) = z + s'$ for some $s' \in D$. Hence for any $x$ sufficiently close to $x_0$, we have, by the triangle inequality:

$|f(x) - f(x_0)| = |f(x - x_0)| = |f(r - s)| = |f(r) - f(s)| = |r' - s'| \leq 2r$.

Shrink the radius $r$ gives the desired result.

shark
  • 789