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.
Best Answer
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.