Connected nowhere dense set with positive measure

Question

everyone. Forgive me for bad English. Can you please give me some clue how to construct a connected space or plane curve, that has positive measure, but nowhere dense?

Answer 1

If $C$ is a fat Cantor set with $0\in C$ then $([0,1]\times C)\cup(\{0\}\times[0,1])$ is nowhere dense, connected and has positive measure.

Getting a curve seems harder. A fat Sierpinski carpet, as suggested, is fine, except it's not immediately clear to me that it's actually a curve. Here's a construction I understand: you can start with a space-filling curve and modify it a bit so it becomes nowhere dense but still has positive measure.

First note that when I say "interval" below I'm referring to a subinterval of $[0,1]$; similarly "squares" will be subsets of $[0,1]^2$.

A dyadic interval is an interval of the form $$I=[j2^{-n},(j+1)2^{-n}]$$with $j,n\in\Bbb N$. A dyadic square is a square $Q=I\times J$, where $I$ and $J$ are dyadic intervals of the same length. If $I$ is an interval and $Q$ is a square then $|I|$ and $|Q|$ will denote the length of $I$ and the length of a side of $Q$, respectively.

Let's say a 4-adic interval is a dyadic interval $I$ with $|I|=4^{-n}$.

One of the standard constructions of a space-filling curve, I think Hilbert's, maybe it's Peano, gives a continuous surjection $\gamma:[0,1]\to[0,1]^2$with the following convenient property: If $I$ is a 4-adic interval of length $4^{-n}$ then $\gamma(I)=Q$ is a dyadiic square with $|Q|=2^{-n}$; further, if $I=[a,b]$ then $\gamma(a),\gamma(b)\in\partial Q$. Note that it follows that $\gamma^{-1}(Q^o)\subset I^o$, because if $I'$ is another 4-adic interval of the same length and $Q'=\gamma(I')$ then $Q$ and $Q'$ have disjoint interiors. (Because, being dyadic squares of the same size, they either have disjoint interiors or are equal, and if $Q'=Q$ there are not enough dyadic squares of the same size left to cover $[0,1]^2$.)

Now let $(Q_k)$ be a dense collection of disjoint dyadic squares with $$\sum|Q_k|^2<1.$$Construct $\gamma'$ as follows:

Say $\gamma'(t)=\gamma(t)$ for $t\in[0,1]\setminus\bigcup I_k$. Now if $I_k=[a_k,b_k]$, define $\gamma'$ on $I_k$ by saying that $\gamma'|_{I_k}$ is a polygonal path in $\partial Q_k$ from $\gamma(a_k)$ to $\gamma(b_k)$.

It's clear that $\gamma'$ is nowhere dense and has positive measure. The only thing that is perhaps not clear is the fact that $\gamma'$ is continuous.

Say $\gamma_n$ is the curve constructed by modifying $\gamma$ on $I_1,\dots,I_n$ as above, leaving $I_k$ alone for $k>n$. It's clear that $\gamma_n$ is continuous, and the fact that $|Q_k|\to0$ shows that $\gamma_n\to\gamma'$ uniformly.