set of arbitrary positive measure conformally equivalent to unit disk Show that for any  $\epsilon$ > 0, there is a dense subset of $\mathbb{C}$ with measure less than $\epsilon$ and which is conformally equivalent to the unit disc.
To make a dense set that has measure small, I was thinking about a disconnected set. But such a set cannot equivalent to unit disck which is connected. What can I do now?
 A: A "starlike" set about $0$ is a set $E\subset \mathbb {C}$ such that $z \in E, t\in [0,1] \implies tz \in E.$ Easily verified: If $\{E_a\}$ is a collection of starlike sets about $0,$, then $\cup E_a$ is starlike about $0.$
Suppose $U$ is open and starlike about $0.$ Then $U$ is simply connected. Proof: Let $\gamma$ be a closed path in $U.$ Take the simplest homotopy you can think of to shrink $\gamma$ to $0$ within $U.$
Suppose $t \in [0,2\pi).$ Let $\epsilon > 0.$ Then the ray $\{re^{it}: r\in [0,\infty)\}$ is contained in an open set $U,$ starlike about $0,$ such that the area of $U$ is less than $\epsilon.$ This is not hard: WLOG $t =0.$ Consider $\{x+iy: |y| < (\epsilon/3) e^{-|x|}\}.$
To finish: Let $\epsilon > 0.$ Let $\{t_n\}$ be dense in $[0,2\pi).$ Let $U_n$ be an open set, starlike about $0,$ that contains the ray $ \{re^{it_n}: r\in [0,\infty)\},$ such that the area of $U_n$ is less than $\epsilon/2^n.$ Then $\cup U_n$ has area $<\epsilon,$ is starlike, hence is simply connected, and is dense. By the Riemann mapping theorem, we're done.
