# Open sets do not have measure zero

I want to show that if $a\lt b$, then $(a,b)$ is not of measure zero.

My idea was to show that any interval covering $(a,b)$ is such that the sum of the lengths of the intervals is always greater than $b-a$.
So I tried induction. Suppose $n=1$, where $n$ is the number of intervals covering $(a,b)$. Then the interval consists of a single set of the form, say, $(c,d)$ such that $(a,b)\subset (c,d)$. If this is the case then obviously, we would have $d-c \gt b-a$, where I have assumed that $c\lt a$ and $b \lt d$.

Now my questions: I am a little bit lost as to how to show it for $n+1$.

Am I even approaching it correctly?

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@Austin: Yes, you start with the usual length on intervals and then you associate outer measure. The tricky bit is to show that you can't cover an interval $(a,b)$ by a family of intervals $I_n$ such that $\sum |I_n| \lt b - a$. See also this MO post by Gowers – t.b. Sep 9 '11 at 5:59
@Theo I agree. I mostly wanted to match up terminology. In any event it is good to mention these things. – Dylan Moreland Sep 9 '11 at 6:11
@All: It is Lebesgue measure on $\mathbb{R}$ – Joe Sep 9 '11 at 6:13

First of all, your open set $(a,b)$ contains a closed interval $[c,d]$ with $c \lt d$. By its definition Lebesgue outer measure is monotone, hence it suffices to show that $\mu^{\ast}[c,d] \gt 0$.

I agree with William that induction won't help much. leo's argument is fine, but I prefer a more informal presentation of the argument to see what is going on.

Every cover of $[c,d]$ contains a finite subcover by compactness and dropping superfluous intervals only reduces the total length of the cover, so we may assume that we deal with finitely many intervals to begin with.

So, let $[c,d] \subset I_1 \cup \cdots \cup I_n$. Since $c$ is covered, we must have an interval $(a_1,b_1)$ in our family $\{I_1,\ldots,I_n\}$ with $c \in (a_1,b_1)$, or, equivalently, $a_1 \lt c \lt b_1$. If $b_1 \lt d$ then $b_1$ is covered by another interval $(a_2,b_2)$ in the family $\{I_1,\ldots,I_n\}$, so $a_2 \lt b_1 \lt b_2$. If $b_2 \lt d$ then $b_2$ is covered$\ldots$

Since we started with a finite family of intervals that covers $[c,d]$ we must at some point arrive at an interval $(a_j,b_j)$ with $a_j \lt d \lt b_j$ and we stop there.

In this way we produce a sequence $(a_1,b_1), (a_2,b_2), \ldots, (a_j,b_j)$ of intervals covering $[c,d]$ such that $a_{i+1} \lt b_{i}$ for $i = 1,\ldots,j-1$ and $a_1 \lt c$ as well as $d \lt b_j$. Now we can estimate \begin{align*} \sum_{i=1}^{n} |I_i| & \geq \sum_{i=1}^{j} (b_i - a_i) =(b_j - a_j) + (b_{j-1} - a_{j-1}) + \cdots + (b_1 - a_1) \\ &= b_j + \underbrace{(b_{j-1} - a_j)}_{\geq 0} + \underbrace{(b_{j-2} - a_{j-1})}_{\geq 0} + \cdots + \underbrace{(b_1 - a_2)}_{\geq 0} - a_1 \\ &\geq b_j - a_1 \geq d-c. \end{align*} This works with an arbitrary finite family of intervals $\{I_1, \ldots,I_n\}$ covering $[c,d]$. Hence this shows that Lebesgue outer measure of $[c,d]$ is at least $\mu^\ast [c,d] \geq d-c$ and since it is clear that it is at most $d-c$, we have $\mu^\ast [c,d] = d-c$.

Finally, we should combine the above with what Carl said, namely that the outer measure of $[a,b]$ is equal to the outer measure of $(a,b)$ and we're done.

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Here's a plan. As Theo suggests, let's take a closed interval $[c, d]$ with $c < d$ contained in $(a, b)$. I claim that there does not exist a cover $\{(a_i, b_i)\}_1^\infty$ of $(a, b)$ such that $\sum b_i - a_i < d - c$.

Note that such a cover would also cover the compact set $[c, d]$. Thus a finite number of the $(a_i, b_i)$ can cover $[c, d]$, and I claim that by throwing out further sets and relabeling we may assume that we have a cover $\{(a_i, b_i)\}_1^n$ of $[c, d]$ in which no element contains another, and such that $b_i \in (a_{i + 1}, b_{i + 1})$ for $i = 1, \ldots, n - 1$.

Having justified all this, show that $$\sum_{i = 1}^n (b_i - a_i) \geq d - c.$$

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You beat me to it! Very nice hint... – t.b. Sep 9 '11 at 6:38
@Theo The dreaded blue "you just wasted five minutes" bar strikes again! You should post yours, if you still have it; I have no issues with deleting this one. – Dylan Moreland Sep 9 '11 at 6:48
Yes... Just leave yours, I'll keep mine and I'll post it if there should be the need of further help. Thanks for the offer, though, and have a good night's rest! – t.b. Sep 9 '11 at 6:57

I'll assume that you use Lebesgue outer measure. Let $m^*$ the outer measure. To prove that the measure of a set isn't $0$, is enough show that exist a $\epsilon_0>0$ such that for any countable covering $\bigcup (a_{k},b_k)$, $\sum_{k=1}^\infty b_k - a_{k}\geq \epsilon_0$.

Take $\epsilon_0=b-a$ and any covering $\bigcup (a_{k},b_k)$. Let $\epsilon>0$, arbitrary. For all $k\in \mathbb{N}$, let $$I_k=(a_k-\epsilon/2^k,b_k).$$ Then $(a_{k},b_k)\subset I_k$ and $$\ell(I_k)=b_k-a_k+\frac{\epsilon}{2^k}.$$ Pick $N\in \mathbb{N}$ such that for $n\geq N$, $$[a+1/2n,b-1/2n]\subset (a,b).$$ Therefore $$[a+1/2n,b-1/2n]\subset (a,b)\subseteq\bigcup (a_{k},b_k)\subset\bigcup I_k.$$ Since the $I_k$ are open sets and cover the compact $[a+1/2n,b-1/2n]$, there is a $M\in\mathbb{N}$ such that (by reidexing if is necessary) $$[a+1/2n,b-1/2n]\subset \bigcup_{k=1}^M I_k.$$ Therefore \begin{align*} \ell([a+1/2n,b-1/2n])&\leq \sum_{k=1}^M \ell(I_k)\\ b-a -\frac{1}{n}&< \sum_{k=1}^M b_k-a_k + \frac{\epsilon}{2^k}\\ &=\sum_{k=1}^M b_k-a_k + \sum_{k=1}^M \frac{\epsilon}{2^k}\\ &\leq \sum_{k=1}^M b_k-a_k + \sum_{k=1}^\infty \frac{\epsilon}{2^k}\\ &= \sum_{k=1}^M b_k-a_k + \epsilon,\end{align*} where $\ell(I)$ denotes the length of a finite interval $I$. Since $\epsilon>0$ is arbitrary $$b-a\leq \sum_{k=1}^M b_k-a_k + \frac{1}{n}.$$ Since the last inequality is true for all $n\geq N$, $$b-a\leq \sum_{k=1}^M b_k-a_k\leq\sum_{k=1}^\infty b_k-a_k.$$ Therefore $$\sum_{k=1}^\infty b_k - a_{k}\geq b-a$$ for any covering $\bigcup (a_k,b_k)$.

Edit: I edit the post because often the outer measure $m^*$ of $A$ is defined as $$m^*(A):=\inf\left\{\sum_{k\in \Delta} \ell(I_k):\, \Delta \text{ countable and } \{I_k\}_{k\in\Delta} \text{ is a covering of } A\right\}.$$

We don't know that the outer measure of a interval of endpoints $a<b$ is $b-a$. This proves it as Carl Mummert point. But we already know that if $I\subset \bigcup I_k$ (this union countable) then $$\ell(I)\leq \sum \ell(I_k),$$ for $I$ finite.

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This is the usual argument. It's worth noticing that this not only shows that the interval has positive outer measure, it shows that the outer measure of $(a,b)$ is at least $b-a$. Because it is trivial to make a covering that has size exactly $b-a$, this means that the outer measure of an interval is exactly its length. – Carl Mummert Sep 9 '11 at 11:17

Note that $m^*[a,b] = m^*(a,b)$, so you can work with the closed interval instead, which is already compact.

Proof in terms of elementary properties of outer measure: $m^*(a,b) \leq m^*[a,b]$ by monotonicity. Also $m^*[a,b] \leq m^*(a,b) + m^*\{a,b\}$ by subadditivity, and $m^*\{a,b\} = 0$, so $m^*[a,b] \leq m^*(a,b)$.

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I am assuming you are using Lebesgue measure on $\mathbb{R}$.

Induction will not prove your result. Even if you manage to use induction successfully, you only would have proved : for all $n$, every covering of $(a,b)$ by $n$ intervals has the property that the sum of the measure is greater than $b - a$, you have not proved this result for an infinite collection of intervals that cover $(a,b)$ but may not have any finite sub collection which covers (note open intervals are not compact).

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Probably Lebesgue outer measure is meant... Anyway, you have monotonicity by definition and $(a,b) \supset [c,d]$ and it suffices to show that $[c,d]$ has positive measure and your point about non-compactness seems moot. – t.b. Sep 9 '11 at 5:57
I was addressing the method of using induction. I was trying to emphasize that induction only proves for all $n$ but there are infinite collections where no finite subset covers in which case proving this result for all $n$ would not even work. Montonicity certainly would give a correct and more concise proof. – William Sep 9 '11 at 6:02
@William: If induction will not, what will? – Joe Sep 9 '11 at 6:12
@Joe: the usual proof uses compactness to reduce down from the general case (an infinite covering) to the case of a finite covering. Then you can use induction or some other method to cover the case of a finite covering. But induction on its own won't handle the general case. – Carl Mummert Sep 9 '11 at 11:19
@Carl: Thanks.... – Joe Sep 9 '11 at 13:12