We construct a bounded open set $A$ in $\mathbf{R}$ such that $BdA$ does not have measure zero. The rational numbers in the open interval $(0,1)$ are countable; let us arrange them in sequence $q_{1},q_{2},\cdots$. Let $0<a<1$ be fixed. For each $i$, choose an open interval $(a_{i},b_{i})$ of length less than $\frac{a}{2^{i}}$ that contains $q_{i}$ and contained in $(0,1)$. Let $A$ be the following open set of $\mathbf{R}$:
$$A=(a_{1},b_{1})\cup(a_{2},b_{2})\cup\cdots=I_{1}\cup I_{2}\cup\cdots$$
In Munkres' Analysis on Manifolds, page 113 Example 1, he proves $BdA$ is not of measure zero. I can understand his proof, but I meet some difficulties when I think this problem in another way.
We know that $BdI_{k}=\{a_{k},b_{k}\}$ is of measure zero (in $\mathbf{R}$). So, $\bigcup_{k=1}^{\infty}BdI_{k}$ is of measure zero.
Since $Bd(a_{1},b_{1})\cup(a_{2},b_{2})\cup\cdots\cup(a_{n},b_{n})\subset\bigcup_{k=1}^{n}BdI_{k}$, we have $BdA\subset\bigcup_{k=1}^{\infty}BdI_{k}$. As a result, we have $BdA$ is of measure zero. But, in fact, $BdA$ is not of measure zero. Where do I go wrong?