Proving the Lebesgue measure is $\sigma$-finite for any dimension

Question

I understand how the Lebesgue measure on the real line is $\sigma$-finite, but I don't understand how to prove that the lebesgue measure is $\sigma$-finite for any dimension. Here I'm using the definition of $\sigma$-finite as follows;

A measure $\mu$ on a measurable space ($X$,$\Sigma$) is $\sigma$-finite if there exists a sequence of measurable sets $E_1$,$E_2$,... $\in$ $\Sigma$ such that $X$ = $\cup^{\infty}_{k=1}$$E_k$ and $\mu (E_k)$ < $\infty$ for every $k \geq 1$.

Thank you to anyone who could explain this to me!

Answer 1

For any $n$, \begin{align*} {\bf{R}}^{n}=\bigcup_{(N_{1},...,N_{n})\in{\bf{Z}}^{n}}\left[N_{1},N_{1}+1\right)\times\cdots\times\left[N_{n},N_{n}+1\right) \end{align*} and $|\left[N_{1},N_{1}+1\right)\times\cdots\times\left[N_{n},N_{n}+1\right)|=1<\infty$ for each $(N_{1},...,N_{n})\in{\bf{Z}}^{n}$.

Answer 2

The space $\mathbb{R}^n$ is just the countable union of $n$-dimensional cubes centered at the origin, each of which has finite measure (since each is a bounded cube). Explicitly, $$ \mathbb{R}^n = \bigcup_{i = 1}^\infty [-i, i]^n. $$ Then the measure of each $E_i = [-i, i]^n$ is finite, and your whole space is the countable union of the $E_i$.