Measure of Interval is Length

Theorem

Let $I$ be a real interval whose endpoints are $a$ and $b$.

Then $I$ is Lebesgue measurable, and the value of the measure is the length of the interval $b - a$.

Proof

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Let $L \subset \R$ be a real interval.

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

Let $\ds \set {I_n}_{n \mathop = 1}^\infty$ be a set of open real intervals satisfying:

$\ds L \subseteq \bigcup_{n \mathop = 1}^\infty I_n$

Case 1: $L$ is open and finite

The theorem follows from the definition of Lebesgue measure.

One can construct:

$I_n = \begin {cases} L & : n = 1 \\ \O & : n \ne 1 \end {cases}$

which yields the sum:

$\ds \sum \map l {I_n} = b - a$

This sum could not be any less because then:

$\ds a + \sum \map l {I_n} < b$

Hence:

$\map m L = b - a$

$\Box$

Case 2: $L$ is closed and finite

The open interval:

$\openint {a - \epsilon} {b + \epsilon}$

contains $\closedint a b$ for each positive $\epsilon$.

So:

$\map m L \le l \closedint {a - \epsilon} {b + \epsilon} = b - a + 2 \epsilon$

Since this is true for any positive $\epsilon$, we have:

$\map m L \le b - a$

Now it must be shown that:

$\map m L \ge b - a$

which will demonstrate:

$\map m L = b - a$

By the Heine-Borel Theorem, any set of open intervals covering $\closedint a b$ contains a finite subcover.

The sum of the lengths of the finite subcover is no greater than the sum of the lengths of the infinite cover.

Therefore it will suffice to show that $\ds \sum \map l {I_n} \ge b - a$ only for finite covers.

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

Call this set $\openint {a_1} {b_1}$.

Necessarily:

$a_1 < a < b_1$

If $b_1 \le b$, then:

$b_1 \in \closedint a b$

So there must exist a set in $\set {I_n}$ containing $b_1$.

Let this set be $\openint {a_2} {b_2}$.

Continuing in this fashion, construct a (finite) sequence of intervals:

$\openint {a_1} {b_1}, \openint {a_2} {b_2}, \ldots, \openint {a_k} {b_k}$

Since $\set {I_n}$ is finite, this process must terminate at some interval $\openint {a_k} {b_k}$.

But this process can only terminate if $b \in \openint {a_k} {b_k}$.

Hence:

$\ds \sum \map l {I_n} \ge \sum_{i \mathop = 1}^k \paren {b_i - a_i} = b_k - a_1 - \sum_{j \mathop = 1}^{k - 1} \paren {a_{j + 1} - b_j} > b_k - a_1$

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

But $b_k > b$ and $a_1 < a$, and so:

$b_k - a_1 > b - a$

Hence

$\ds \sum \map l {I_n} \ge b - a$

$\Box$

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:

$\map l J > \map l L - \epsilon$

Hence:

$\map l L - \epsilon < \map l J = \map m J \le \map m L \le \map m {\map c L} = \map l {\map c L} = \map l L$

where $\map c L$ is the closure of $L$.

Thus for each positive $\epsilon$:

$\map l L - \epsilon < \map m L \le \map l L$

and so:

$\map m L = \map l L = b - a$

$\Box$

Case 4: Infinite Intervals

If $L$ is infinite, either $a$ or $b$ is $\pm \infty$.

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

Hence $\map m L \ge \Delta$ for arbitrarily large $\Delta$, and so:

$\map m L = \infty$

$\blacksquare$