How to use Lebesgue measure formula I'm struggling to see how to use the Lebesgue measure formula. So for example given any subset $[a,b]$ of the reals, what is it's "length" according to the formula:
$\lambda(A)=\inf({\sum|b_i-a_i|:\text{A subset of the union}\ [a_i, b_i] })$
 A: Let me reformulate: 
$\lambda(A)=\inf S$ where $s\in S\subseteq[0,\infty]$ if and only if a countable family $\langle a_i,b_i\rangle_{i\in I}$ in $\mathbb R^2$  exists with $a_i\leq b_i$, $A\subseteq\bigcup_{i\in I}[a_i,b_i]$ and $s=\sum_{i\in I}(b_i-a_i)$.
If $A=[a,b]$ for $a,b\in\mathbb R$ with $a\leq b$ then it is obvious that $b-a\in S$ so that $\inf S\leq b-a$.
It remains to prove that  $\inf S\geq b-a$ or equivalently that $s\geq b-a$ for every $s\in S$.
Let $s\in S$ and let it correspond with countable family $\langle a_i,b_i\rangle_{i\in I}$ so that $[a,b]\subseteq\bigcup_{i\in I}[a_i,b_i]$
With induction to $n$ it can be proved that for every interval $[p,q]\subset\mathbb R$ and every positive integer $n$ we have:$$[p,q]\subseteq\bigcup_{i=1}^n[p_i,q_i]\implies q-p\leq\sum_{i=1}^n(q_i-p_i)\tag1$$
This is enough for the case where $I$ is finite.
Now if $I$ is not finite then for a fixed $\epsilon>0$ we can construct $\epsilon_i>0$ with $\sum_{i\in I}\epsilon_i<\epsilon$.
We have $[a,b]\subseteq\bigcup_{i\in I}(a_i-\epsilon_i,b+\epsilon_i)$ and $[a,b]$ is compact, so there is a finite $I_0\subset I$ with: $$[a,b]\subseteq\bigcup_{i\in I_o}(a_i-\epsilon_i,b+\epsilon_i)\subseteq\bigcup_{i\in I_o}[a_i-\epsilon_i,b+\epsilon_i]$$
Then with $(1)$ we find: $$b-a\leq\sum_{i\in I_0}(b_i-a_i+2\epsilon_i)\leq\sum_{i\in I}(b_i-a_i)+2\epsilon=s+2\epsilon$$
This is true for every $\epsilon>0$ so eventually we arrive at: $$b-a\leq\sum_{i\in I}(b_i-a_i)=s$$
