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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.

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  • $\begingroup$ "...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 $\mathbb R$. $\endgroup$ – Bungo Sep 9 at 7:25
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$(x_n,y) \in X_n$ 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 $(x_n-\frac {\epsilon} {2^{k}},x_n+\frac {\epsilon} {2^{k}}) \times (x_k-1,x_k+1)$.

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  • $\begingroup$ 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. $\endgroup$ – Bungo Sep 9 at 7:34
  • $\begingroup$ @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. $\endgroup$ – Kabo Murphy Sep 9 at 7:38
  • $\begingroup$ OK, fair enough. $\endgroup$ – Bungo Sep 9 at 7:41
  • $\begingroup$ @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? $\endgroup$ – 21understanding Sep 9 at 9:01
  • $\begingroup$ Yes. If you take small intervals around all the rationals their union may still be a small set in measure theoretic sense, though this union is dense. $\endgroup$ – Kabo Murphy Sep 9 at 9:03

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