Let $A$ be an absorbing set of $\mathbb{R}^2$, that is, $\bigcup_{r>0} \bigcap_{|t|>r} tA = \mathbb{R}^2$.
Is it necessary that $A$ has a positive Lebesgue measure?
Or can there be such an $A$ with zero Lebesgue measure?
Or can there be such an $A$ with no Lebesgue measure at all? A non-measurable absorbing set of $\mathbb{R}^2$.
$A$ may not be measurable. For instance, consider $V$ a Vitali-like subset of $(-\pi,\pi]$, identify $\Bbb R^2=\Bbb C$ and call $A=\{z\in\Bbb C\,:\, \lvert z\rvert\le1\land \operatorname{arg}z\in V\}\cup\{z\in\Bbb C\,:\, \lvert z\rvert\le2\land \operatorname{arg}z\notin V\}$.
Or, in a simpler way, consider the unit ball $B$ united with your favourite non-measurable subset of $\Bbb R^2$ which is disjoint from $B$.
However, if $A$ is measurable, then its measure is positive. In fact, notice that $$\Bbb R^2=\bigcup_{r\in \Bbb Q^+}\bigcap_{\lvert t\rvert\in(r,\infty)\cap \Bbb Q}tA$$
and the RHS is a null set if $A$ is. To prove the identity, notice that, for all $r\in\Bbb R$, $\bigcap_{\lvert t\rvert\in(r,\infty)\cap \Bbb Q}tA\supseteq\bigcap_{\lvert t\rvert>r}tA$, and therefore $\bigcup_{r\in \Bbb Q^+}\bigcap_{\lvert t\rvert\in(r,\infty)\cap \Bbb Q}tA\supseteq \bigcup_{r\in \Bbb Q^+}\bigcap_{\lvert t\rvert>r}tA$. Moreover, the map $x\mapsto \bigcap_{\lvert t\rvert>x}tA$ (for $x\in\Bbb R$) is weakly increasing, therefore $\bigcup_{r\in\Bbb Q^+}\bigcap_{\lvert t\rvert>r}tA\supseteq \bigcap_{\lvert t\rvert>x}tA$ for all $x>0$ by considering some $r\in \Bbb Q$ larger than $x$. Thus $$\bigcup_{r\in \Bbb Q^+}\bigcap_{\lvert t\rvert\in(r,\infty)\cap \Bbb Q}tA\supseteq\bigcup_{r\in\Bbb Q^+}\bigcap_{\lvert t\rvert>r}tA\supseteq\bigcup_{x>0}\bigcap_{\lvert t\rvert>x}tA=\Bbb R^2$$