# Lebesgue measure of $[a,b]$ in $\mathbb R ^n$

If we know that the outer Lebesgue measure of the interval $(a,b)$ in $\mathbb R ^n$ is $\displaystyle{ \prod_{j=1}^{n} (b_j - a_j)}$ find the outer Lebesgue measure of the closed interval $[a,b]$.

Here it is what I did:

$a= (a_1, a_2, \cdots ,a_n)$ and $b=(b_1 ,b_2, \cdots ,b_n)$

$$\prod_{j=1}^{n} (b_j - a_j) = m^{*}((a,b)) \leq m^{*}([a,b]) \leq m^{*} ( (a-\epsilon , b+ \epsilon)) \quad \forall \epsilon >0$$

But it is $\displaystyle{ m^{*} ( (a-\epsilon , b+ \epsilon)) = \prod_{j=1}^{n} (b_j - a_j +2 \epsilon) = \prod_{j=1}^{n} (b_j - a_j) + A(\epsilon) }$ where $A(\epsilon)$ are the terms of the product with $\epsilon$. Since the above holds for all $\epsilon > 0$ the coclusion follows.

Can you help me write the above in a more elegant way?

Thanks in advance!

edit: Actually I am wondering about the part that I wrote about the terms A(ϵ)

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## 1 Answer

Actually it works very nicely. It just depends on the continuity of the map $$(x_1, x_2, \cdots , x_n) \mapsto \prod_{k=1}^n x_k.$$

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O.K Actually I am wondering about the part that I wrote about the terms $A(\epsilon)$. – passenger May 27 '12 at 19:01
The continuity guarantees that $\lim_{\epsilon\to 0} A(\epsilon) = 0.$ – ncmathsadist May 27 '12 at 19:19
O.K I see! Thank you very much for your reply! – passenger May 27 '12 at 19:25