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So far, we have only constructed the Lebesgue measure on intervals. Now if $\{I_k\}$ is a countable cover of the reals by intervals which overlap at one point at most, we say that $S\subset \mathbb R$ is measurable iff $S \cap I_k $ is measurable in $I_k$ for all $k$ and set $$m(S)=\sum_{k=0}^{\infty}m_{I_k}(S\cap I_k)$$ I have shown that the values coincide for any two covers, knowing that $S$ is measurable for all elements in both covers. How do we show that if S is measurable according to one cover, then it is measurable according to another? (Then showing that the family of measurable sets is a $\sigma$-algebra and additivity of the measure should pose little problem.)

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For two covers $\{I_k\}$ and $\{J_l\}$, use their intersections $\{I_k\cap J_l\}$, so that $I_k=\bigcup_l (I_k\cap J_l)$ and $J_l=\bigcup_k (I_k\cap J_l)$. –  Berci Feb 8 '13 at 16:09
    
What do you mean by "$S\cap I_k$ measurable in $I_k$"? You say you've defined measure only for intervals but the intersection isn't always an interval –  leo Feb 9 '13 at 19:27
    
We have constructed the Lebesgue measure on an interval. Not only subintevals are measurable. –  Student Feb 10 '13 at 10:06
    
@Berci Your suggestion is sound to show that the measure of a subset does not depend on the cover, given it is measurable according to both covers. If you read the question carefully, you will notice that this is not the issue. –  Student Feb 10 '13 at 10:16
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You should show that if $I\subset J$ and $S\cap I$ is measurable in $I$ then $S\cap I$ is measurable in $J$. Then given any two covers you can reduce them to a common subcover.

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Yes, that is the result I'm trying to get. How can I show that $S\cap I$ is measurable in $J$. –  Student Feb 8 '13 at 16:28
    
How is defined measurability on an interval? –  Emanuele Paolini Feb 8 '13 at 16:46
    
$S \subset I$ is measurable $\Leftrightarrow I \setminus S$ is measurable $\Leftrightarrow m_\star(S)=m^\star(S) \Leftrightarrow (\forall \epsilon >0)(\exists U \quad open)(\exists K \quad compact)(K \subset S \subset U \land m(U\setminus K)<\epsilon)$ –  Student Feb 8 '13 at 17:29
    
As you can see the definition does not depend on the interval –  Emanuele Paolini Feb 8 '13 at 17:59
    
I am still not convinced. If we take the second definition with the inner and outer measures, the supremum will get smaller, and the infimum will get bigger, hence we can only establish an inequality. If we take the third, what is to guarantee that such open and compact sets exist not only in I but in J as well? –  Student Feb 10 '13 at 10:12
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