Can a compact connected set in $\mathbb{R}^n$ with empty interior have positive Lebesgue measure? Is there a compact connected subset if $\mathbb{R}^n$, with empty interior but positive Lebesgue measure?
 A: Here is an example in $\mathbb R^2.$
First, define a "fat Cantor set" in $\mathbb R^1$ like this: Instead of deleting the middle third of $[0,1],$ we delete (in this present example) the middle quarter.
Then delete the middle quarter of each of the two remaining intervals.
Then delete the middle quarter of each of the four remaining intervals.
And so on.
Our fat Cantor set will contain every point that never gets deleted.
Its measure is $\displaystyle 1 - \frac 1 4 - \frac 2 {4^2} - \frac 4 {4^3} - \frac 8 {4^4} - \cdots = 1 - \frac{1/4}{1-(1/2)} = \frac 1 2.$
Its interior is empty because within every open interval some points get deleted.
Now let this fat Cantor set be on the $x$-axis, and look at the set of all points between any point on that fat Cantor set and $(0,1).$ That's a connected set in $\mathbb R^2$ that has empty interior and positive measure.
A: YES.
First construct a "Cantor like set" with positive measure $C$ - otherwise known as fat Cantor set. Then define the product $C\times [0,1]\subset \mathbb R^2$ and finally to make it connected define
$$
T=C\times [0,1]\cup [0,1]\times\{0\}.
$$
What is a Cantor like set?
Take $C_0=[0,1]$, then $C_1=[0,1]\setminus [a_1^0,b_1^0]$, then $C_2=C_1\setminus([a_1^1,b_1^1]\cup [a_2^1,b_2^1])$, so that $[a_1^1,b_1^1]\subset (0,a_1)$ and $[a_2^1,b_2^1]\subset (b_1,1)$ and continue this process as in the construction of the Cantor set, but so that $\sum m([a_i^j,b_i^j])<1$.
