Let $\mathbb{R}^{n}$ be equipped with the $\sigma$-algebra of Lebesgue measurable sets, and Lebesgue measure.

(a) Given $f \in L^{\infty}\left(\mathbb{R}^{n}\right), g \in L^{1}\left(\mathbb{R}^{n}\right)$, define the convolution $f \star g$, and show that it is a bounded, continuous function. [You may use without proof continuity of translation on $L^{p}\left(\mathbb{R}^{n}\right)$ for $\left.1 \leqslant p<\infty .\right]$

Suppose $A \subset \mathbb{R}^{n}$ is a measurable set with $0<|A|<\infty$ where $|A|$ denotes the Lebesgue measure of $A$. By considering the convolution of $f(x)=\mathbb{1}_{A}(x)$ and $g(x)=\mathbb{1}_{A}(-x)$, or otherwise, show that the set $A-A=\{x-y: x, y \in A\}$ contains an open neighbourhood of 0 . Does this still hold if $|A|=\infty$ ?

(b) Suppose that $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is a measurable function satisfying

$$f(x+y)=f(x)+f(y), \quad \text { for all } x, y \in \mathbb{R}^{n}$$

Let $B_{r}=\left\{y \in \mathbb{R}^{m}:|y|<r\right\}$. Show that for any $\epsilon>0$ :

(i) $f^{-1}\left(B_{\epsilon}\right)-f^{-1}\left(B_{\epsilon}\right) \subset f^{-1}\left(B_{2 \epsilon}\right)$,

(ii) $f^{-1}\left(B_{k \epsilon}\right)=k f^{-1}\left(B_{\epsilon}\right)$ for all $k \in \mathbb{N}$, where for $\lambda>0$ and $A \subset \mathbb{R}^{n}, \lambda A$ denotes the set $\{\lambda x: x \in A\}$.

Show that $f$ is continuous at 0 and hence deduce that $f$ is continuous everywhere.