# Outer measure of $\mathbb{Q}\times \mathbb{R}$

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.

• "...then $y$ is in a small open interval $(x_k-\frac{1}{2^{k+1}},x_k+\frac{1}{2^{k+1}})$ for some $k$..." - This is not true. The sum of the lengths of these intervals is $\sum_{k=1}^{\infty}1/2^k = 1$, so there's no way they can cover all of $\mathbb R$. – Bungo Sep 9 at 7:25

$$(x_n,y) \in X_n$$ 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 $$(x_n-\frac {\epsilon} {2^{k}},x_n+\frac {\epsilon} {2^{k}}) \times (x_k-1,x_k+1)$$.
• For each fixed $n$, your proposed covering has the same positive measure, so taking the union over all $n$ you'll get infinity. The width of the first interval in the product needs to depend on both $n$ and $k$. I think replacing $\epsilon / 2^k$ with $\epsilon / 2^{k+n}$ would do the trick. – Bungo Sep 9 at 7:34
• @Bungo OP has already mentioned that he is using sigma subadditivity so it is enough to show that $\lamda^{*}(X_n)=0$ for each fixed $n$. From that point on $n$ is fixed and there is no need to take union over $n$ in the inequalities. – Kabo Murphy Sep 9 at 7:38
• @KaviRamaMurthy I see. Thank you, sir. I think I use the density of the real line in the wrong way. I shouldn't be able to guarantee that any irrational is in such small $I_k$ with rational end points. Correct? – 21understanding Sep 9 at 9:01