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Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of $R^{n}$

Statement

Let $n \geq 1$ , assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $E \subseteq R^{n}$ . Then $E$ is Lebesgue measurable (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ) if and only if each of the following four conditions holds, and the four are equivalent to one another.

Open excess. For every real $ε > 0$ there is an open $U \supseteq E$ with $λ_{n}^{*} (U ∖ E) < ε$ .
$G_{δ}$ minus null. There are a $G_{δ}$ set $G$ and a set $Z$ with $λ_{n}^{*} (Z) = 0$ and $E = G ∖ Z$ ( $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion).
Closed deficit. For every real $ε > 0$ there is a closed $F \subseteq E$ with $λ_{n}^{*} (E ∖ F) < ε$ .
$F_{σ}$ plus null. There are an $F_{σ}$ set $H$ and a set $W$ with $λ_{n}^{*} (W) = 0$ and $E = H \cup W$ .

Each condition is stated for sets of infinite measure as well as finite ones, which is why the excess and the deficit are measured by the outer measure of a difference rather than by a difference of measures.

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, and a subset $E \subseteq R^{n}$ .

[L1]

Assuming countable choice, for every Lebesgue measurable $E$ and every real $ε > 0$ there is an open $U$ with $E \subseteq U$ and $λ_{n}^{*} (U ∖ E) < ε$ (For a Lebesgue measurable set and every positive $ε$ there is an open superset whose difference from it has outer measure below $ε$ ).

[L2]

Assuming countable choice, a set admitting open supersets of arbitrarily small outer excess is $G ∖ Z$ with $G$ a $G_{δ}$ containing it and $λ_{n}^{*} (Z) = 0$ , and is Lebesgue measurable (A subset of $R^{n}$ with open supersets of arbitrarily small excess is Lebesgue measurable).

[L3]

Assuming countable choice, $L (R^{n})$ is a sigma-algebra, $λ_{n}$ is a complete measure on it, and every $S$ with $λ_{n}^{*} (S) = 0$ is Lebesgue measurable of measure $0$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[L4]

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

[L5]

Assuming countable choice, $λ_{n}^{*}$ is an outer measure on $R^{n}$ , hence monotone (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

[L6]

Assuming countable choice, $λ_{n}^{*} (E) = \inf {λ_{n} (U) : U$ 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).

[F1]

$A$ is a $G_{δ}$ set of $X$ when there is a sequence $(V_{n})_{n \in N}$ of open subsets of $X$ with $A = ⋂_{n \in N} V_{n}$ , and an $F_{σ}$ set of $X$ when there is a sequence $(F_{n})_{n \in N}$ of closed subsets with $A = ⋃_{n \in N} F_{n}$ ( $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion).

[F2]

A subset $F \subseteq X$ is closed in $(X, d)$ if its complement $X ∖ F$ is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and the product topology on $R^{n}$ is the metric topology of $d_{2}$ (A subset of $R^{n}$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology, claim 1).

[F3]

For every real $ε > 0$ there is a natural number $k \geq 1$ with $1 / k < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[F4]

The Axiom of Countable Choice says that for every family $(X_{n})_{n \in N}$ of nonempty sets indexed by $N$ there is a function $f$ with domain $N$ such that $f (n) \in X_{n}$ for every $n \in N$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1

Measurability implies condition 1, which is the cited lemma on the open excess of a measurable set.

1.2

Condition 1 implies condition 2, which is the first clause of the cited lemma on small open excess.

1.3

Condition 2 implies measurability: $G$ is a countable intersection of open sets, hence Borel and measurable; $Z$ has outer measure $0$ , hence is measurable; so $E = G ∖ Z$ is measurable.

L3L4F1

1.4

Condition 4 implies measurability, by the same argument read for unions: $H$ is a countable union of closed sets, hence Borel and measurable, $W$ is measurable because $λ_{n}^{*} (W) = 0$ , and $E = H \cup W$ is measurable.

L3L4F1

2.1

Measurability implies condition 3: the complement $R^{n} ∖ E$ is measurable, so for a real $ε > 0$ step 1.1 supplies an open $U \supseteq R^{n} ∖ E$ with $λ_{n}^{*} (U ∖ (R^{n} ∖ E)) < ε$ ; then $F : = R^{n} ∖ U$ is closed, $F \subseteq E$ , and $E ∖ F = E \cap U = U ∖ (R^{n} ∖ E)$ , so $λ_{n}^{*} (E ∖ F) < ε$ .

step 1.1L3F2

3.1

Condition 3 implies condition 4: for each $m \in N$ the family of closed $F \subseteq E$ with $λ_{n}^{*} (E ∖ F) < 1 / (m + 1)$ is nonempty, so countable choice selects such an $F_{m}$ ; then $H : = ⋃_{m} F_{m}$ is an $F_{σ}$ set with $H \subseteq E$ , and $W : = E ∖ H \subseteq E ∖ F_{m}$ gives $λ_{n}^{*} (W) \leq 1 / (m + 1)$ for every $m$ , hence $λ_{n}^{*} (W) = 0$ and $E = H \cup W$ .

step 2.1L5F1F3F4

4.1

The implications of steps 1.1, 1.2 and 1.3 close the cycle between measurability and conditions 1 and 2, and those of steps 2.1, 3.1 and 1.4 close the cycle between measurability and conditions 3 and 4; so all five statements are equivalent, and outer regularity is what stands behind the open sets produced in step 1.1.

step 1.1step 1.2step 1.3step 1.4step 2.1step 3.1L6∎

Depends on

Used by

Dependency tree · two levels

73 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), Exercises 1.2.7 and 1.2.19 (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)
E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Theorem 1.6 (standard reference, not scraped)

Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of RnRn