Measure of Interval is Length

Theorem
An interval of the real numbers is Lebesgue measurable and the value of the measure is the length of the interval.

Proof
Let $L \subset \R$ be an interval of real numbers.

Then $L$ has two distinct endpoints, $a$ and $b$.

Let $\displaystyle \left\{{I_n}\right\}_{n=1}^\infty$ be a set of open intervals satisfying $\displaystyle L \subseteq \bigcup_{n=1}^\infty I_n$.

Case 1: $L$ is open and finite
The theorem follows from the definition of Lebesgue measure.

One can construct:
 * $I_n = \begin{cases} L & : n = 1 \\ \varnothing & : n \ne 1 \end{cases}$

which yields a $\sum l (I_n) = b - a$.

This sum could not be any less because then $a + \sum l(I_n) < b$.

Hence $m(L) = b-a$

Case 2: $L$ is closed and finite
The open interval $(a-\epsilon . . b+\epsilon)$ contains $[a. . b]$ for each positive $\epsilon$, so $m(L) \leq l (a - \epsilon . . b + \epsilon) = b-a+2\epsilon$.

Since this is true for any positive $\epsilon$, we have $m(L) \leq b-a$.

Now it must be shown that $m(L) \geq b-a$; this would demonstrate $m(L) = b-a$.

By the Heine-Borel Theorem, any collection of open intervals covering $[a. . b]$ contains a finite subcover.

Since the sum of the lengths of the finite subcover is no greater than the sum of the lengths of the infinite cover, it will suffice to show that $\sum l(I_n) \geq b-a$ only for finite covers.

Since $a \in \bigcup I_n$, there must be a set in $\left\{{I_n}\right\}$ containing $a$.

Call this set $(a_1,b_1)$.

Necessarily, $a_1 < a < b_1$.

If $b_1 \leq b$, then $b_1 \in [a. . b]$.

So there must exist a set in $\left\{{I_n}\right\}$ containing $b_1$; let this set be $(a_2 . . b_2)$.

Continuing in this fashion, construct a series of intervals $(a_1 . . b_1), (a_2 . . b_2), \ldots, (a_k . . b_k)$.

Since $\left\{{I_n}\right\}$ is finite, this process must terminate at some interval $(a_k . . b_k)$.

But this process can only terminate if $b \in (a_k . . b_k)$.

Hence:


 * $\displaystyle \sum l(I_n) \geq \sum_{i=1}^k (b_i-a_i) = b_k - a_1 - \sum_{j=1}^{k-1} (a_{j+1}-b_j) > b_k - a_1$

since $a_i > b_{i-1}$.

But $b_k > b$ and $a_1 < a$, and so $b_k - a_1 > b-a$.

Hence $\displaystyle \sum l(I_n) \geq b-a$.

Case 3: $L$ is a finite interval
Regardless of whether the set in question is open, closed, or possibly neither, given $\epsilon >0$, there is a closed interval $J \subset L$ such that $l(J) > l(L) - \epsilon$.

Hence:
 * $l(L) - \epsilon < l(J) = m(J) \leq m(L) \leq m(c(L)) = l(c(L)) = l(L)$

where $c(L)$ is the closure of $L$.

Thus for each positive $\epsilon$, $l(L)-\epsilon < m(L) \leq l(L)$ and so $m(L) = l(L) = b - a$.

Case 4: Infinite Intervals
If $L$ is infinite, either $a$ or $b$ is $+/- \infty$.

Given any real number $\Delta$, there is a closed interval $J \subset L$ with $l(J)= \Delta$.

Hence $m(L) \geq \Delta$ for arbitrarily large $\Delta$, and so $m(L)=\infty$.