Lebesgue Measure of $\mathbb{R} \times \{0\}$

Question

I know this is probably obvious, and I know the answer is that it is (Lebesgue) measure zero, but I'm having a hard time wrapping my head around it. Looking for an intuitive explanation.

Question: What is the measure of: $\mathbb{R} \times \{0\}$, where $\times$ is the Cartesian product.

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Answer 1

$$m(\mathbb{R}\times\{0\})=m\left(\bigcup_{n=1}^\infty[-n,n]\times\{0\}\right)\leq\sum_{n=1}^\infty m([-n,n]\times \{0\})=\sum_{n=1}^\infty (2n)\cdot 0=0.$$

Intuitive explanation: A line has no area.

Answer 2

Let $\lambda$ denote the Lebesgue measure. Then since $\mathbb{R}$ and $\{ 0\}$ are measurable sets, if $\mu=\lambda \times \lambda$ is the product measure $$ \mu(\mathbb{R} \times \{0\})= \lambda(\mathbb{R})\cdot\lambda(\{0\})=\infty \cdot 0 = 0 $$

Answer 3

Take the rectangles $[n,n+1]\times[-\varepsilon/2^{|n|},\varepsilon/2^{|n|}]$ for $n\in\mathbb Z$ and fixed $\varepsilon>0$. Then their area is $2\varepsilon/2^n$, and $\mathbb R\times\{0\}$ is contained in the union of all these rectangles. Summing up the areas gives $6\varepsilon$, so the (two)-dimensional Lebesgue measure of $\mathbb R\times\{0\}$ must be $\leq 6\varepsilon$. Since $\varepsilon$ was arbitrary, the result follows.