It is needed to find a measure which delivers length. Recall P

(S)

denotes the set of all subsets of
S.

Theorem 5.7.1There exists a function m : P

(ℝ)

→

[0,∞ ]

which satisfies the followingproperties.

If A ⊆ B, then 0 ≤ m

(A )

≤ m

(B)

,m

(∅)

= 0.

m

(∪∞ Ai)
k=1

≤∑_{i=1}^{∞}m

(Ai )

m

([a,b])

= b − a = m

((a,b))

.

Proof: First it is necessary to define the function m. This is contained in the following
definition.

Definition 5.7.2For A ⊆ ℝ,

{ ∞ }
m (A) = inf ∑ (bi − ai) : A ⊆ ∪∞ (ai,bi)
i=1 i=1

In words, you look at all coverings of A with open intervals. For each of these open coverings, you add
the lengths of the individual open intervals and you take the infimum of all such numbers
obtained.

Then 1.) is obvious because if a countable collection of open intervals covers B, then it also covers
A. Thus the set of numbers obtained for B is smaller than the set of numbers for A. Why is
m

(∅)

= 0? Then ∅⊆

(a − δ,a + δ)

and so m

(∅)

≤ 2δ for every δ > 0. Letting δ → 0, it follows that
m

(∅)

= 0.

Consider 2.). If any m

(Ai)

= ∞, there is nothing to prove. The assertion simply is ∞≤∞. Assume
then that m

(Ai )

< ∞ for all i. Then for each m ∈ ℕ there exists a countable set of open intervals,