What is missing in this outer measure proof of interval?

Question

Let $m^*(I)$ be the outer measure of interval $I$, and take $|I|$ as the interval's length. In my textbook, it's trivial that $m^*(I) \leq |I|$ since $m^*(I)$ is the infimum. What's not trivial is the other direction $m^*(I) \geq |I|$, which needs the Heine-Borel theorem, and the proof goes like:

Given $\epsilon>0$, since $m^*(I)$ is the infimum, we'll have:

$$m^*(I)+\epsilon\geq \sum_{k \in O}|k|.$$

where $O$ is an open cover of $I$. Since $I$ is compact, we can find a finite open cover (using Heine-Borel) $O_K$, with $K=|O_K|$, s.t.:

$$\sum_{k \in O_K}^Kl(k)\geq |I|$$

thus, we can make $\epsilon \rightarrow 0^+$ in the equation $m^*(I) \geq |I| - \epsilon$ to get $m^*(I) \geq |I|$

Now, I came up with this simpler proof that does not use Heine-Borel, but am sure there's something missing...

Given $\epsilon>0$, since $m^*(I)$ is the infimum, we'll have:

$$m^*(I)+\epsilon\geq \sum_{k \in O}|k|$$

where $O$ is any open cover of $I$. Using the elements of $O$, we can form a new set of intervals $J$, where elements $j \in J$ are formed from the endpoints of intervals in $O$. For instance, if there are two intervals $k_1=(a,b)$ and $k_2=(c,d)$ in $O$, where $a < c < b < d$, we form three new intervals in $J$, namely $[a-\epsilon,c+\epsilon]$,$[c-\epsilon,b+\epsilon]$, and $[b-\epsilon,d+\epsilon]$. This could be done for all elements $k \in O$. If there are three intervals $k_1=(a,b)$,$k_2=(c,d)$,$k_3=(e,f)$ with $a < c < e < b < d < f$, we'll have $[a-\epsilon,c+\epsilon]$,$[c-\epsilon,e+\epsilon]$,$[e-\epsilon,b+\epsilon]$,$[b-\epsilon,d+\epsilon]$,$[d-\epsilon,f+\epsilon]$.. and so on and so forth.

If any intervals $\in O$ overlap, we 'chop up' the overlaps into almost disjoint intervals. We'll have a large set of closed almost disjoint intervals $j \in J$. So $J$ covers $I$ and the overlaps can be minimised, and we can then have (abusing notation for $\epsilon$):

$$l(I) \leq \sum_{j \in J} |j| \leq \sum_{k \in O} |k| \leq m^*(I) + \epsilon$$

we then take $\epsilon \rightarrow 0^+$ as before (by minimizing the overlaps among the $j \in J$).

I sense that there's something wrong with my simpler proof (since it does not use Heine-Borel) but I couldn't figure it out... Any help?

P.S. Can this have something do with what's happening in the Cantor set, i.e. I can form an infinite number of intervals with rational endpoints that cover $I$ in this way, i.e. say I first divide $I$ into two halves (with $|I|=1$). So these intervals would be $[a,b]$ and $[c,d]$ where $a$,$b$,$c$,$d$ are rational. I then divide again each of $[a,b]$ and $[c,d]$ into two, and do this up to infinity, so the resulting interval length is $[\frac{1}{2}]^n$ -- which becomes measure zero at infinity ala Cantor set, i.e. the intervals that supposedly cover $I$ would no longer be intervals at the limit.

Answer 1

If a cover is infinite, it could be much more complicated than what you are imagining. That is the point of using Heine-Borel to reduce to a finite cover.

Suppose we use the definition of $m^*$ on the rational numbers instead of on the real numbers. Length of an interval makes sense in the rationals. Why does your argument not apply in that case? In the rationals, in fact, $m^*(I) = 0$ for all intervals $I$.

Answer 2

To begin with your method will fail for the cover $ \mathit{O} = \{ I_1,I_2,\dots \} $ if $ I_1 \subset I_2 \subset \cdots $. A first approach to amending the proof might be to first remove intervals that contain other intervals and then run the procedure. However, after we run the procedure we might again have produced intervals that contain other intervals.

The other thing we can do is modify the algorithm a bit to be the following: For a given pair $ (I,J) \in \mathit{O}\times\mathit{O} $, replace $ J $ by $ J \setminus (a+\delta,b-\delta), \; \delta>0 $.

Let $ I = (a,b) $ and let $ \mathit{O} $ be an open cover of $ I $. By intersecting the intervals of $ \mathit{O} $ with $ (a-\varepsilon/4,b+\varepsilon/4) $ we can always ensure that $ m^*(\bigcup \mathit{O}) \leq |I| + \varepsilon/2 $. Thus we have that $$ \sum_{J\in \mathit{O}} |J| \leq m^*(\bigcup \mathit{O}) + \sum_{J,J'\in \mathit{O}} |J\cap J'| \leq |I| + \varepsilon/2 + \sum_{J,J'\in \mathit{O}} |J\cap J'|, $$ so it remains to show that we can modify $ \mathit{O} $ so that $ \sum_{J,J'\in \mathit{O}} |J\cap J'| \leq \varepsilon/2 $. To do this let $ (J_1,J_1'),(J_2,J_2'),\dots $ be an ordering of $ \mathit{O} \times \mathit{O} $. In step $ i $ we run the procedure with $ \delta = \varepsilon2^{-3-i} $ and amend all the pairs with the new intervals to the list. Thus in the end we will have that $$ \sum_{J,J'\in \mathit{O}} |J\cap J'| = \sum_{i=1}^{\infty} |J_i\cap J_i'| \leq \sum_{i=1}^{\infty} \varepsilon 2\cdot 2^{-3-i} = \varepsilon/2. $$ This completes the proof which, assuming it works, is now considerably more complicated.