# lebesgue measure of $\{ (x,y,z) \in \mathbb R ^3 : x \in \mathbb R, 0 \leq y \leq 10, z \in \mathbb Z \}$

Find the lebesgue measure of the set: $$\Bigl\{ (x,y,z) \in \mathbb R ^3 : x \in \mathbb R, \quad 0 \leq y \leq 10, \quad z \in \mathbb Z \Bigr\}$$

I think is a null set but for some reason I have stuck and I can't write down a complete solution.

Any help?

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Well, technically the Lebesgue measure assigns a number (not a set) to your set, so you must be thinking of measure zero". What do you know about product measures / the Lebesgue measure of a countable set in $\mathbb{R}?$ – user17794 Jul 2 '12 at 23:03
@TimDuff: The Lebesgue measure of a countable set in $\mathbb R$ is zero. What do you mean by product measures? – passenger Jul 2 '12 at 23:05

$$M:=\{(x,y,z) \in \mathbb{R}^3:\ x \in \mathbb{R}, \ 0 \le y \le 10, \ z \in \mathbb{Z}\}= \bigcup_{i,j \in \mathbb{Z}}M_{i,j},$$ where each $M_{ij}:=[i,i+1]\times[0,10]\times\{k\}$ has measure $0$ since for every $\epsilon>0$ one has $M_{ij} \subset M_{ij}^\epsilon:=[i,i+1]\times[0,10]\times[-\epsilon/20,\epsilon/20]$ and $|M_{ij}^\epsilon|=\epsilon$. Therefore $M$ is a countable union of null sets, and thus $M$ is a set of measure $0$.

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Thank you very much for your solution! – passenger Jul 2 '12 at 23:10

You can use countable additivity of Lebesgue measure along the $x$-axis and $z$-axis. Specifically your set is the disjoint union $\cup_{n,z\in\mathbb Z}\{(x,y,z)| x\in (n,n+1], y\in[0,10]\},$ each of these having measure $0.$

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Thank you for your answer! – passenger Jul 2 '12 at 23:10
You're welcome! – Andrew Jul 2 '12 at 23:31

Let's call the set $X$. The Lebesgue measure $m$ is inner and outer-regular, so in order to show $m(X)=0$ we need only show that any compact subset $C$ of $X$ has measure $0$, which in turn we can do by showing that for any $\epsilon>0$ there is an open set $U\supset C$ with $m(U)<\epsilon$.

Since $C$ is compact, it must be bounded, thus we have upper bounds for $|x|$ and $|z|$. Let's denote these $M_x$ and $M_z$. Note that $$C\subseteq C'=\bigcup\limits_{z=-M_z}^{M_z} [-M_x,M_x]\times [0,10]\times \{z\}$$ and since $m$ is monotonic $m(C)\leq m(C')$. To see that $m(C')=0$, let $$U_n=\bigcup\limits_{z=-M_z}^{M_z} (-M_x-1,M_x+1)\times (-1,11)\times (z-1/n,z+1/n)$$ which is open and a finite union of rectangles, so its measure is easy to calculate. Specifically, $m(U_n)=(2M_x+2)\times12\times 2/n$, which goes to $0$ as $n\to \infty$. Thus $m(C')=0$ so $m(C)=0$ for any compact set $C\subseteq X$, hence $m(X)=0$.

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