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For bounded subsets of $R$ , Lebesgue measurability is equivalent to equality of inner and outer measure

Statement

Assume the Axiom of Countable Choice. Let $E \subseteq R$ be bounded. Then $E$ is Lebesgue measurable if and only if

$λ_{*} (E) = λ^{*} (E),$

where $λ_{*} (E)$ is the inner measure of Lebesgue inner measure on the real line and $λ^{*} (E)$ is the Lebesgue outer measure.

Facts & Assumptions

Given: The Axiom of Countable Choice and a bounded set $E \subseteq R$ .

[F1]

$λ_{*} (E) = \sup {λ (K) : K \subseteq E and K is compact}$ (Lebesgue inner measure on the real line).

[L1]

Assuming countable choice, a subset of $R$ is Lebesgue measurable if and only if for every real $ε > 0$ there is an open $U \supseteq E$ with $λ^{*} (U ∖ E) < ε$ , and if and only if for every real $ε > 0$ there is a closed $F \subseteq E$ with $λ^{*} (E ∖ F) < ε$ (Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of $R^{n}$ , clauses 1 and 3).

[L2]

Assuming countable choice, $λ^{*} (E) = \inf {λ (U) : U \subseteq R open and E \subseteq U}$ (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[L3]

A subset of $R$ is compact if and only if it is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded).

[L4]

Assuming countable choice, every Borel subset of $R$ is Lebesgue measurable (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable).

[L5]

Assuming countable choice, $(R, L (R), λ)$ is a complete measure space (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

Proof

technique · direct

1.1F1L1L3L4L5

For the forward implication, assume $E$ is Lebesgue measurable and fix a real $ε > 0$ . By [L1] choose a closed $F \subseteq E$ with $λ^{*} (E ∖ F) < ε$ ; since $E$ is bounded, so is $F$ , hence [L3] makes $F$ compact. The set $F$ is Borel and therefore measurable by [L4], so $λ (F) \leq λ_{*} (E) \leq λ^{*} (E)$ by [F1]; and $E = F ⊔ (E ∖ F)$ with both pieces measurable by [L4] and [L5], so $λ^{*} (E) = λ (E) = λ (F) + λ (E ∖ F) < λ_{*} (E) + ε$ . As $ε > 0$ was arbitrary, $λ^{*} (E) \leq λ_{*} (E)$ ; together with $λ_{*} (E) \leq λ^{*} (E)$ from [F1], this gives $λ_{*} (E) = λ^{*} (E)$ .

1.2F1L2L3L4L5choose

For the converse implication, assume $λ_{*} (E) = λ^{*} (E)$ and fix a real $ε > 0$ . By [F1] choose a compact $K \subseteq E$ with $λ (K) > λ_{*} (E) - ε / 2$ , and by [L2] choose an open $U \supseteq E$ with $λ (U) < λ^{*} (E) + ε / 2$ . The compact set $K$ is Borel by [L3], hence measurable by [L4], and $K \subseteq E \subseteq U$ ; therefore $U ∖ E \subseteq U ∖ K$ , so [L5] gives $λ^{*} (U ∖ E) \leq λ (U ∖ K) = λ (U) - λ (K) < λ^{*} (E) - λ_{*} (E) + ε = ε$ .

2.1step 1.2L1∎

Step 1.2 gives open supersets of arbitrarily small outer excess, so [L1] makes $E$ Lebesgue measurable. This is the reverse implication.

Remarks

The boundedness hypothesis is used only to turn the closed set $F$ of step 1.1 into a compact set, which is what allows the inner measure to see it.
On an unbounded measurable set the equality $λ_{*} (E) = λ^{*} (E)$ can be vacuous, since both sides may be $+ \infty$ .

Depends on

Used by

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.2.18 (standard reference, not scraped)
John K. Hunter, Measure Theory (UC Davis lecture notes), Theorems 2.24, 2.25 and 2.27 (standard reference, not scraped)

For bounded subsets of RR, Lebesgue measurability is equivalent to equality of inner and outer measure