Is there a compact connected subset if $\mathbb{R}^n$, with empty interior but positive Lebesgue measure?
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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 $\sum m([a_i^j,b_i^j])<1$.
Yiorgos S. Smyrlis
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So it is connected but not path-connected am I right? – Yanko Jan 31 '18 at 12:35
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2It is path connected. Join any two points by getting down to $[0,1]\times {0}$, then right/left to the desired $x$-value, then up again. – Arnaud Mortier Jan 31 '18 at 12:41
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@ArnaudMortier Ok now It's clear, that's why he added that $[0,1]\times {0}$ – Yanko Jan 31 '18 at 12:43
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Yes, otherwise it wouldn't even be connected. This is called a Smith-Volterra-Cantor set, or fat Cantor set. – Arnaud Mortier Jan 31 '18 at 12:43
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2So it is a comb like structure with tooth protruding from the points of Cantor set? – Prajwal Kansakar Jan 31 '18 at 12:52
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Yes, but not any Cantor set if you want to ensure positive measure in the end. – Arnaud Mortier Jan 31 '18 at 13:41
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This set is even polygonally connected. In fact, it's an $L_2$ set in the sense of Horn/Valentine's 1949 paper Some properties of $L$-sets in the plane (see also Bruckner/Bruckner's 1962 paper On $L_n$ sets, the Hausdorff metric, and connectedness). – Dave L. Renfro Jan 31 '18 at 14:15
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@ArnaudMortier Oh yes, I see now! – Prajwal Kansakar Jan 31 '18 at 14:39
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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.
