Given $X=\{(x,y)\mid x\in \mathbb{Q}, y\in \mathbb{R}\}$. I want to show that $\lambda^*(X)=0$, where $\lambda^*$ is the Lebesgue outer measure.
My attempt:
Let the enumeration of $\mathbb{Q}$ be $\{x_1,x_2,x_3,…\}$. Hence, we have
$$
X=\bigcup_{n=1}^{\infty}X_n,
$$
where $X_n=\{(x_n,y)\mid y\in \mathbb{R}\}$. We show that for every $n$, we have $\lambda^*(X_n)=0$, so that by the countably subadditivity of Lebesgue outer measure, we have $\lambda^*(X)=0$.
Let $n\geq 1$. Take $\epsilon>0$. For each $k$, consider the open cell
$$
I_k=\left(x_n-\frac{\epsilon}{4},x_n+\frac{\epsilon}{4}\right)\times \left(x_k-\frac{1}{2^{k+1}},x_k+\frac{1}{2^{k+1}}\right).
$$
If $(x_n,y)\in X_n$, then $y$ is in a small open interval $\big(x_k-\frac{1}{2^{k+1}},x_k+\frac{1}{2^{k+1}}\big)$ for some $k$, so that $(x_n,y)\in I_k$. This means that $X_n \subseteq I_1 \cup I_2 \cup I_3 \cup \ldots$.
Observe that, if $l(I_n)$ is the length of $I_n$, we have
$$
\sum_{k=1}^{\infty} l(I_k)=\sum_{k=1}^{\infty}2(\frac{\epsilon}{4})2(\frac{1}{2^{k+1}})=\frac{\epsilon}{2}<\epsilon.
$$ Thus, $\lambda^*(X_n)=0$ for every $n$, and hence $\lambda^*(X)=0$.
Is this proof correct? I am not sure about the reasoning on $(x_n,y)$ part.
Thank you.
Best Answer
$(x_n,y) \in X_n$ does not tell you anything about $y$. It can be any real number. So your assertion that $y \in (x_k-\frac 1 {2^{k+1}},x_k+\frac 1 {2^{k+1}})$ for some $k$ is not true. The union of these intervals does not exhaust the real line because their union has finite measure.
For a correct proof you can use the intervals $(x_n-\frac {\epsilon} {2^{k}},x_n+\frac {\epsilon} {2^{k}}) \times (x_k-1,x_k+1)$.