OK, maybe this won't be interested to seasoned people, but anyway. The question comes from Stein/Shakarchi and it asks for an example of an open set whose boundary has non-zero measure. At first I thought this was trivial, but it helped clarify for me an important point that measure is relative to the dimension of the ambient Euclidean space (we're obviously not dealing with abstract measure here).

Anyhow, I just spent about three hours proving a bunch of properties about "fat" Cantor sets, and the example is as follows:

Let $\displaystyle \hat{C}$ be a fat Cantor set so constructed such that $\displaystyle \mu(\hat{C}) = 1 - \sum_{k=1}^{\infty}2^{k-1}l_{k} = 1 - \mu(O) > 0$. Here $\displaystyle l_{k}$ denotes the length corresponding to the $\displaystyle 2^{k-1}$ open segments removed at the kth stage (note that it varies according to k). The sum which appears is proved to be the measure of the complement of $\displaystyle \hat{C}$ and by various other points the entire expression gives the measure of $\displaystyle \hat{C}$ (this was all proved in a separate problem, which is interesting in itself since it allows us to construct Cantor-like sets which retain many rather paradoxical properties, but now with positive measure).

The hint in the text book suggests that we construct the open set by considering the union of the open segments deleted at the odd (or even) stages, and consider the closure of this set (which presumably has positive measure). This seems a little confusing, and it's not clear at all to me why you would want to consider the odd/even stages.

So here is my spin of the solution:

Just let $\displaystyle O$ be the union of all open segments removed in the construction of $\displaystyle \hat{C}$. I proved in a separate exercise that $\displaystyle \forall\;x\in\hat{C}\;\exists\;\{x_{n}\}_{n=1}^{\i nfty}\subset\hat{C^{c}}=O$ such that $\displaystyle x_{n}\to x$ and each $\displaystyle x_n\in I_{n}$ such that $\displaystyle I_{n}\to 0$. This I will take as a lemma now.

Applying the above lemma, we see that $\displaystyle \forall\;x\in\hat{C}$ there exists a sequence contained in $\displaystyle O$ which converges to $\displaystyle x$. This means that $\displaystyle x$ is a limit point of $\displaystyle O$ (there is an easy theorem that shows these two concepts are actually equivalent in any metric space). But $\displaystyle x\notin O$, so $\displaystyle x\in\partial O$. And since the measure of $\displaystyle \hat{C}$ is non-zero, we see that the boundary of $\displaystyle O$ has non-zero measure as well.

Is this correct, or is it really necessary to consider the odd/even stages correctly? I know a lot of statements in my "proof" need to be proved themselves, but rest assured, I've proved them (tediously) in a separately typed up LaTeX paper.