Lebesgue outer measure satisfies $\lambda^{*}([a,b]) \leq b-a$

Question

Aaagain, I fail to understand the trivial:

Using compactness argument it is straightforward to show: $$\lambda^{*}([a,b]) \geq b-a$$ And everything is OK.

But, regarding $\lambda^{*}([a,b]) \leq b-a$, Cohn says:

For any closed bounded interval $[a,b] \subset \mathbb{R}$ it is easy to see that $\lambda^{*}([a,b]) \leq b-a$ (cover $[a,b]$ with sequences of open intervals in which the first interval is barely larger than $[a,b]$, and the sum of the lengths of the other intervals is very small).

From this hint I should be able to arrive at $\lambda^{*}([a,b]) < b-a + \epsilon$, for arbitrary $\epsilon >0$, but I don't understand how..

Help?

@user2345215 the question is about $\lambda^{*}([a,b]) \leq b-a$ using what Cohn suggested. — user76568, Mar 07 '14 at 17:05
Everything you may wish for is in this question and the comments below: http://math.stackexchange.com/questions/174517/on-lebesgue-outer-measure-of-an-interval — Lord Soth, Mar 07 '14 at 17:11
@LordSoth Thank you! You think I should delete this question? — user76568, Mar 07 '14 at 17:13
@Dror I think deleting a question is not a good idea, we can mark it as duplicate though. — Lord Soth, Mar 07 '14 at 17:14
@Integral I'd say that makes two of us, but I'm not going to, because "dumb" isn't it. Try "still clueless" :-). — user76568, Mar 07 '14 at 17:18
I'm not sure what's your exact definition of outer measure, but if it's something like infimum of sum of lengths of covering intervals with length defined as b-a, then doesn't it follow from the covering by that exact interval (or almost exact if it can't be closed)? — user2345215, Mar 07 '14 at 17:22
@Dror: Why don't you cover it by $(a-\frac\varepsilon2,b+\frac\varepsilon2)$ with length $b-a+\varepsilon$ then? — user2345215, Mar 07 '14 at 17:35
@user2345215 It needs to be covered by an infinite sequence of open intervals — user76568, Mar 07 '14 at 17:37
@Dror: Empty interval is open. (and if you can't use that - that's an arbitrary restriction, I wouldn't want a definition which disallows it, that's just plain stupid). Nevertheless, I'll post the answer with nonempty intervals as well. — user2345215, Mar 07 '14 at 17:38
@user2345215 calm down.. And perhaps think twice before you call something "stupid" — user76568, Mar 07 '14 at 17:41

score 1 · Accepted Answer · answered Mar 07 '14 at 17:50

The simplest solution is to cover your your interval $[a,b]$ with just $1$ interval $(a-\frac\varepsilon2,b+\frac\varepsilon2)$ of length $b-a+\varepsilon.$ That's why it's trivial.

If you definition is needlessly restrictive and requires infinitely many open intervals, set empty sets for the remaining intervals.

If your definition of an interval is needlessly restrictive, you can cover your set like this $$\left(a-\frac\varepsilon4,b+\frac\varepsilon4\right)\cup\bigcup_{n=3}^\infty\left(-\frac\varepsilon{2^n},\frac\varepsilon{2^n}\right)$$ of total length $b-a+\frac\varepsilon2+\sum\limits_{n=2}^\infty\frac\varepsilon{2^n}=b-a+\frac\varepsilon2+\frac\varepsilon2=b-a+\varepsilon$.

Maybe now you see why is it stupid to not allow what I suggest.

Thanks. But note that It's not my definition, and I will not call It stupid, because I do not yet know if these restrictions are in fact useful or not. My ""guess"" is that they are useful. — user76568, Mar 07 '14 at 18:00

Lebesgue outer measure satisfies $\lambda^{*}([a,b]) \leq b-a$

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