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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 1

up vote 1 down vote accepted

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

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