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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!

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    $\begingroup$ Do you want to prove it's $\sigma-$finite? It must be $\sigma$-additive to be a measure. $\endgroup$ – saulspatz Mar 10 '18 at 19:27
  • $\begingroup$ Yes, sorry I do. Edited now! @saulspatz $\endgroup$ – Giraffes4thewin Mar 10 '18 at 19:38
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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 $(N_{1},...,N_{n})\in{\bf{Z}}^{n}$.

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  • $\begingroup$ Thank you, so where did you get the $N+1$ from and how did you figure out that the measure is 1? $\endgroup$ – Giraffes4thewin Mar 12 '18 at 15:19
  • $\begingroup$ About the measure, simply because for boxes, one splits: $|[N_{1},N_{1}+1)\times\cdots\times[N_{n},N_{n}+1)|=|[N_{1},N_{1}+1)|\times\cdots\times|[N_{n},N_{n}+1)|=1\times\cdots\times 1=1$. $\endgroup$ – user284331 Mar 12 '18 at 15:30
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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 $E_i$.

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