Exercise 1.2.3 (The outer measure axioms)

Show that

(i)
(Empty set) m∗(∅) = 0.
(ii)
(Monotonicity) If E ⊆ F ⊆ ℝd, then m∗(E) ≤ m∗(F).
(iii)
(Countable subadditivity) If E1,E2,… ⊂ ℝd is a countable sequence of sets, then m∗ (⋃ ⁡ n=1∞En) ≤∑ ⁡ n=1∞m∗(En).

Answers

Notice that by non-negativity of the elementary measure, the Lebesgue outer measure has to remain non-negative as well.

(i)
∅ is contained in any degenerate box { a } d of elementary measure 0; thus, m ∗ ( ∅ ) ≤ 0 . By the previous remark, Lebesgue outer measure is non-negative; thus, it must be zero.
(ii)
Let E ⊆ F ⊆ ℝ d . Since any outer cover of F is also an outer cover of E , we have an obvious relation { ∑ n = 1 ∞ | B n | : B n  are boxes E ⊂ ⋃ n = 1 ∞ B n } ⊇ { ∑ n = 1 ∞ | C n | : C n  are boxes F ⊂ ⋃ n = 1 ∞ C n } .

The infimum of a superset is always smaller than the infimum of any of its subsets; thus,

m ∗ ( E ) ≤ m ∗ ( F ) .

(iii)
Let E 1 , E 2 , … ⊂ ℝ d be a countable sequence of sets; our goal is to demonstrate that inf ⁡ { ∑ n = 1 ∞ | B n | : B n  are boxes ⋃ i = 1 n E i ⊂ ⋃ n = 1 ∞ B n } ≤ ∑ i = 1 n inf ⁡ { ∑ n = 1 ∞ | C n ( i ) | : C n ( i )  are boxes E i ⊂ ⋃ n = 1 ∞ C n ( i ) } ( + 𝜖 )

for every 𝜖 > 0 . By definition of infimum, for 𝜖 > 0 we can find a cover C 1 ( i ) , … of E i such that ∑ ⁡ n = 1 ∞ | C n ( i ) | − m ∗ ( E i ) ≤ 𝜖 ∕ 2 i . Thus, by the countable axiom of choice, we can construct a double infinite sequence

( C ( i , n ) ) i ∈ ℕ , n ∈ ℕ  such that  ∀ ⁡ i ∈ ℕ : E i ⊆ ⋃ n = 1 ∞ C ( i , n )  and  ∑ n = 1 ∞ | C ( i , n ) | ≤ m ∗ ( E i ) + 𝜖 ∕ 2 i .

But we immediately notice that (cf. Tonelli’s theorem for series)

⋃ i ∈ ℕ E i ⊆ ⋃ i ∈ ℕ ⋃ n ∈ ℕ C ( i , n ) = ⋃ ( i , n ) ∈ ℕ 2 C ( i , n ) ⟹ ∑ ( i , n ) ∈ ℕ 2 | C ( i , n ) | ∈ { ∑ n = 1 ∞ | B n | : B n  are boxes ⋃ i = 1 n E i ⊂ ⋃ n = 1 ∞ B n }

which satisfies

m ∗ ( ⋃ i = 1 n E i ) ≤ ∑ ( i , n ) ∈ ℕ 2 | C ( i , n ) | = ∑ i ∈ ℕ ∑ n ∈ ℕ | C ( i , n ) | ≤ ∑ i ∈ ℕ ( m ∗ ( E i ) + 𝜖 ∕ 2 i ) = ∑ i ∈ ℕ m ∗ ( E i ) + 𝜖

as desired.

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2020-05-30 00:00
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First, ∅ is a box of volume 0 ; another proof is by noting that ∅⊂ [0, 1 n ] d for all n, so m∗(∅) ≤ inf ⁡ nm ( [0, 1 n ] d) = 0.

For A ⊂ C we have that the infimum used to obtain m∗(A) is over a larger set that the one used to obtain the infimum for m∗(C).

Let (An) n be a sequence of sets. Fix 𝜀 > 0. For every n let (Bn,k) k be a sequence of boxes such that An ⊂⋃ ⁡ kBn,k and

∑ k |Bn,k| ≤ m∗ (A n) + 𝜀 ⋅ 2−n

Consider the sequence of boxes (Bn,k) n,k. We have that

⋃ nAn ⊂⋃ n,kBn,k

and so

m∗ (⋃ nAn) ≤∑ n,k |Bn,k| ≤∑ nm∗ (A n) + ∑ n𝜀 ⋅ 2−n = ∑ nm∗ (A n) + 2𝜀

Taking 𝜀 → 0 we have countable subadditivity.

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2021-12-04 17:15
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