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?
Thanks for your help.
 A: 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
$$
\sum_{i = 1}^n (b_i - a_i) \geq d - c.
$$
A: 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.
A: 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 $m^*[a,b] \leq m^*(a,b)$. 
A: 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). 
A: 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.
