We define the Lebesgue Outer Measure of an interval $[a,b]$ by

$$\lambda([a,b])= \inf \left\{\sum_{j=1}^\infty |I_j|: [a,b]\subset \bigcup_{j=1}^{\infty} I_j\right\}$$

where $\{I_j\}$ is a sequence of open intervals that cover $[a,b]$, and we define the length of an open interval $|(a,b)|= b-a$. I want to show $\lambda([a,b])= b-a$

The proof of the book that I am using starts with the following:

Let $\epsilon >0$. Because

$$[a,b] \subset \left(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}\right) \cup \bigcup_{n=2}^\infty \left(- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n}\right)$$, we obtain $\lambda([a,b]) < b-a +\epsilon$ which implies $\lambda([a,b])\le b-a$.

I can understand $[a,b] \subset (a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})$, but I do not get the part $\cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ Since $(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})$ already covers $[a,b]$, and I do not think that $\bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ will necessarily cover an arbitrary closed interval $[a,b]$, so I do not really know why you need that union of arbitrary small intervals. (My guessing is that you add the countable union so that it coincides with the definition of Lebesgue outer measure)

In addition, I do not know how can we arrive at the inequality $\lambda([a,b]) < b-a +\epsilon$. I do know that $b-a + \epsilon$ is essentially the length of $(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}) \cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ since I already computed it out. Taking the infimum of the length, we get $b-a$ and this is the outer measure (I think this statement should be wrong though)

Sidenote: I have not officially encountered Lebesgue Measure yet. I encountered this outer measure thing when the book is trying to show function continuous a.e is Riemann-integrable.

I really hope someone can shed some light since I have been puzzled by this the entire day.

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    $\begingroup$ This looks borderline nonsensical. Is this actually the proof in a textbook? Surely you have $[a,b] \subset (a-\epsilon/2, b+\epsilon/2)$ for all $\epsilon>0$, hence $\lambda([a,b])\le b-a+\epsilon$ for all $\epsilon>0$, and hence $\lambda([a,b])\le b-a$. What is the additional stuff for? $\endgroup$ – mjqxxxx Jul 24 '12 at 5:52
  • $\begingroup$ @mjqxxxx this is first part of the proof that yields $\lambda([a,b] \le b-a$. My question is how you yield $\lambda([a,b]) \le b-a + \epsilon$. This might be a trivial step, but I am totally missing it now. $\endgroup$ – Daniel Jul 24 '12 at 6:03
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    $\begingroup$ related $\endgroup$ – leo Jul 24 '12 at 18:36
  • $\begingroup$ @Daniel : proving the obvious fact $\lambda([a,b]) \geq b-a$ is surprisingly challenging. Does anyone know a slick proof? $\endgroup$ – Stefan Smith Dec 9 '13 at 2:23

Apparently the author only wants to use non-empty open intervals (normally, $\emptyset$ is considered an open interval: this follows from the fact that both open sets and intervals are closed under finite intersection).

So because there are infinitely many terms, he must make the size of these intervals small enough so that their lengths sum to $\varepsilon/2$. How they are placed and whether they overlap doesn't matter at all.

To obtain the bound, he is just applying the definition of $\lambda$ with $I_1=(a-\varepsilon/4,b+\varepsilon/4)$ and $I_n=(-\varepsilon/2^{n+1},\varepsilon/2^{n+1})$ for $n\ge 2$. Because this is a covering of $[a,b]$, it appears in the $\inf$ defining $\lambda$: we have $$\lambda([a,b])\le \sum_{j\ge 1} |I_j|=|I_1|+\sum_{n\ge 2} \varepsilon\cdot 2^{-n}=(b-a+\varepsilon/2)+\varepsilon/2$$

This problem could have been avoided altogether by allowing empty intervals in the definition of $\lambda$ with $|\emptyset|=0$: this defines the same $\lambda$, because we can always replace $\emptyset$ with arbitrarily small intervals so that the $\inf$ is unchanged.


It looks like the author want's to formally adhere to the given definition. Adding those intervals does no harm to the fact that $[a,b]\subset (a-\epsilon/4, b+-\epsilon/4)$. Note that the definition is rather challenging regarding the fact that it ignores the fact that some intervals may overlap and their contribution is counted several times, so you cannot do without taking that part of the definition into account. It's correct but does not look too elegant, though.

Also note $\lambda(-\epsilon/2^{n+1},\epsilon/2^{n+1})=\epsilon/2^{n}$ and $\sum_{n\ge 2} \epsilon/2^{n}=\varepsilon/4$.

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    $\begingroup$ I did compute the geometric series by myself, but I still couldn't get to the inequality $\lambda<b-a+\epsilon$, not sure how the intermediate steps work. $\endgroup$ – Daniel Jul 24 '12 at 5:28
  • $\begingroup$ Good observation. $\endgroup$ – copper.hat Jul 24 '12 at 6:21

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