Outer Measure and the Caratheodory Extension Theorem

The test set decomposes as $A=(A\cap E)\cup (A\setminus E)$.

Apply countable subadditivity in [F1] to the two sets in step 1.1 followed by empty sets. This gives ${\mu }^{\ast }(A)\le {\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\setminus E)$, including $0=0+0$ when $A=\mathrm{\varnothing }$ and without subtracting when a value is $+\mathrm{\infty }$.

For every test set $A$, the split by $\mathrm{\varnothing }$ reads ${\mu }^{\ast }(A)=0+{\mu }^{\ast }(A)$, so $\mathrm{\varnothing }$ is measurable; the identity for $X\setminus E$ is the identity for $E$ with its two summands exchanged. Applying [F1] first to $E$ and then, inside each resulting piece, to $F$ splits $A$ into the four cells $A\cap E\cap F$, $A\cap E\setminus F$, $A\cap F\setminus E$, and $A\setminus (E\cup F)$, with their outer measures summing to ${\mu }^{\ast }(A)$.

By subadditivity, ${\mu }^{\ast }(A\cap (E\cup F))$ is at most the sum of the first three cell values from step 1.1, while $A\setminus (E\cup F)$ is the fourth cell; hence ${\mu }^{\ast }(A)\ge {\mu }^{\ast }(A\cap (E\cup F))+{\mu }^{\ast }(A\setminus (E\cup F))$. The reverse inequality is [L1], so $E\cup F$ is measurable and the measurable family is an algebra.

At $n=0$ the finite union is empty and the finite sum is the empty sum $0$, so the formula is ${\mu }^{\ast }(A)=0+{\mu }^{\ast }(A)$; moreover every partial union $U_n:={\bigcup }_{k<n}{E}_k$ is measurable by [L1].

Assume the displayed formula at $n$ and put $R_n:=A\setminus U_n$.

Since $E_n$ is disjoint from $U_n$, [F1] gives ${\mu }^{\ast }(R_n)={\mu }^{\ast }(A\cap E_n)+{\mu }^{\ast }(A\setminus U_{n+1})$. Substituting this equality into the induction hypothesis in step 1.2 gives the formula for $n+1$, including empty pieces and infinite values without cancellation.

Step 1.1 is the base case and step 2.1 proves the successor case, so the formula holds for every $n\in \mathbb{N}$.

Applying [L1] to the first $n$ sets and to $A$ gives ${\mu }^{\ast }(A)\ge {\sum }_{k<n}{\mu }^{\ast }(A\cap E_k)+{\mu }^{\ast }(A\setminus E)$, because $A\setminus E\subseteq A\setminus {\bigcup }_{k<n}{E}_k$ and outer measure is monotone; this includes $n=0$.

Taking the supremum over $n$ in step 1.1 yields ${\mu }^{\ast }(A)\ge {\sum }_k{\mu }^{\ast }(A\cap E_k)+{\mu }^{\ast }(A\setminus E)$. Apply step 1.1 again with the test set $A\cap E$: its remainder outside $E$ is empty, so taking the supremum gives ${\mu }^{\ast }(A\cap E)\ge {\sum }_k{\mu }^{\ast }(A\cap E_k)$ without subtracting an infinite quantity. Countable subadditivity gives the reverse inequality. Substituting this equality into the first bound and using subadditivity on $(A\cap E)\cup (A\setminus E)$ proves the Carathéodory identity, even when the series is $+\mathrm{\infty }$.

Monotonicity gives $0\le {\mu }^{\ast }(S)\le {\mu }^{\ast }(N)=0$, so ${\mu }^{\ast }(S)=0$; for every test set $A$, the same argument gives ${\mu }^{\ast }(A\cap S)=0$.

Since $A\setminus S\subseteq A$, monotonicity gives ${\mu }^{\ast }(A)\ge {\mu }^{\ast }(A\setminus S)={\mu }^{\ast }(A\cap S)+{\mu }^{\ast }(A\setminus S)$ by step 1.1; [L1] supplies the reverse inequality, so $S$ is Carathéodory measurable.

The family ${\mathcal{M}}_{{\mu }^{\ast }}$ is an algebra by [L1] and is closed under countable disjoint unions by [L2], so [L4] makes it a sigma-algebra.

For a pairwise disjoint sequence $(E_k)$ in ${\mathcal{M}}_{{\mu }^{\ast }}$, use $A={\bigcup }_k{E}_k$ in [L2]; then $A\cap E_k=E_k$, so ${\mu }^{\ast }({\bigcup }_k{E}_k)={\sum }_k{\mu }^{\ast }(E_k)$, including infinite values, and the restriction is a measure.

If $N\in {\mathcal{M}}_{{\mu }^{\ast }}$ has restricted measure zero and $S\subseteq N$, then [L3] makes $S$ measurable with outer measure zero; hence the restricted measure is complete.

If some ${\mu }^{\ast }(E_n)=+\mathrm{\infty }$, monotonicity gives the result immediately. Otherwise countable choice and [F1] give measurable hulls $H_n\supseteq E_n$ with ${\mu }^{\ast }(H_n)={\mu }^{\ast }(E_n)$; put $G_n:={\bigcap }_{k\ge n}{H}_k$. Sigma-algebra closure in [L2] makes each $G_n$ measurable, while $G_n\subseteq G_{n+1}$ and $E_n\subseteq G_n\subseteq H_n$, so monotonicity makes ${\mu }^{\ast }(G_n)={\mu }^{\ast }(E_n)$.

Let $G={\bigcup }_n{G}_n$. Since $E_n\subseteq G_n$, one has $E\subseteq G$, while [L1] for the Carathéodory restriction gives ${\mu }^{\ast }(G)={\sup }_n{\mu }^{\ast }(G_n)={\sup }_n{\mu }^{\ast }(E_n)$; hence monotonicity gives ${\mu }^{\ast }(E)\le {\sup }_n{\mu }^{\ast }(E_n)$, and the reverse inequality follows from $E_n\subseteq E$.

The sequence consisting only of empty sets covers $\mathrm{\varnothing }$ at cost $0$, so ${\mu }^{\ast }(\mathrm{\varnothing })=0$; if $E\subseteq F$, every cover of $F$ covers $E$, so taking infima gives ${\mu }^{\ast }(E)\le {\mu }^{\ast }(F)$.

Let $(E_j)$ be a sequence. If some ${\mu }^{\ast }(E_j)=+\mathrm{\infty }$, the desired subadditive inequality is automatic. Otherwise, for $\epsilon >0$, [F1] selects for each $j$ a cover $(C_{jk}{)}_k$ of $E_j$ with cost below ${\mu }^{\ast }(E_j)+\epsilon 2^{-(j+1)}$; [L2] enumerates the doubly indexed family as one sequence covering ${\bigcup }_j{E}_j$, and [L1] computes its cost as at most ${\sum }_j{\mu }^{\ast }(E_j)+\epsilon$, since the displayed geometric error series has partial sums $\epsilon (1-2^{-n})$. Letting $\epsilon$ decrease to $0$ proves countable subadditivity, so with step 1.1 the function is an outer measure.

The algebra law in [F1] gives $\mathrm{\varnothing },X\in {\mathcal{A}}_0$, and the premeasure normalization gives ${\mu }_0(\mathrm{\varnothing })=0$. Thus $\mathcal{C}={\mathcal{A}}_0$ and $p={\mu }_0$ satisfy every hypothesis of [L1].

Applying [L1] to the data in step 1.1 shows that the induced outer set function ${\mu }^{\ast }$ is an outer measure.

Define $B_k:=A\cap (A_k\setminus {\bigcup }_{j<k}{A}_j)$, using the empty preceding union when $k=0$; finite unions, differences, and intersections keep every $B_k$ in ${\mathcal{A}}_0$, and $B_k\subseteq A_k$.

Distinct pieces are disjoint because a point in $B_k$ belongs to no earlier $A_j$; every $x\in A$ lies in some $A_k$, and at its least such index it belongs to $B_k$, so $A={\bigcup }_k{B}_k$.

The sequence $(A,\mathrm{\varnothing },\mathrm{\varnothing },\dots )$ covers $A$ at cost ${\mu }_0(A)$, so ${\mu }^{\ast }(A)\le {\mu }_0(A)$.

For any algebra cover $(A_k)$ of $A$, [L1] gives disjoint $B_k\subseteq A_k$ with union $A$; [F1] gives ${\mu }_0(A)={\sum }_k{\mu }_0(B_k)$, and finite additivity applied to $A_k=B_k\bigsqcup (A_k\setminus B_k)$ gives ${\mu }_0(B_k)\le {\mu }_0(A_k)$, hence ${\mu }_0(A)\le {\sum }_k{\mu }_0(A_k)$.

Since step 1.2 bounds every covering cost below by ${\mu }_0(A)$, [F2] gives ${\mu }_0(A)\le {\mu }^{\ast }(A)$; combining this with step 1.1 proves equality, including infinite values without subtraction.

If ${\mu }^{\ast }(A)=+\mathrm{\infty }$, the reverse Carathéodory inequality is automatic. If ${\mu }^{\ast }(A)<+\mathrm{\infty }$, then for any $\epsilon >0$ choose an algebra cover $(C_k)$ of $A$ with ${\sum }_k{\mu }_0(C_k)<{\mu }^{\ast }(A)+\epsilon$; since $E\in {\mathcal{A}}_0$, the two sequences $(C_k\cap E)$ and $(C_k\setminus E)$ lie in the algebra, cover $A\cap E$ and $A\setminus E$, and [F1] gives ${\mu }_0(C_k)={\mu }_0(C_k\cap E)+{\mu }_0(C_k\setminus E)$.

By the defining infima, step 1.1 gives ${\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\setminus E)<{\mu }^{\ast }(A)+\epsilon$; since this holds for every positive $\epsilon$, the left side is at most ${\mu }^{\ast }(A)$. Countable choice makes ${\mu }^{\ast }$ an outer measure by [L2], so [L1] applies to it and gives the reverse inequality. Thus the Carathéodory identity holds for every $A$.

By [L1], ${\mu }^{\ast }$ is an outer measure, and [L2] makes its restriction to ${\mathcal{M}}_{{\mu }^{\ast }}$ a complete measure.

By [L4], ${\mathcal{A}}_0\subseteq {\mathcal{M}}_{{\mu }^{\ast }}$, and [L3] identifies the restricted values there with ${\mu }_0$.

Since ${\mathcal{M}}_{{\mu }^{\ast }}$ is a sigma-algebra containing ${\mathcal{A}}_0$, [L5] gives $\sigma ({\mathcal{A}}_0)\subseteq {\mathcal{M}}_{{\mu }^{\ast }}$; restricting the measure from step 1.1 to this generated sigma-algebra and using step 1.2 gives the claimed extension, with no sigma-finiteness hypothesis.

If ${\mu }^{\ast }(E)=+\mathrm{\infty }$, put $H=X$. Otherwise, for each $n\in \mathbb{N}$, countable choice and [F2] give an algebra cover $(A_{nk}{)}_k$ of $E$ with cost below ${\mu }^{\ast }(E)+1\mathrm{/}(n+1)$; put $U_n={\bigcup }_k{A}_{nk}$ and $H={\bigcap }_n{U}_n$.

In the finite case each $U_n$ and their intersection $H$ lie in $\sigma ({\mathcal{A}}_0)$; in the infinite case $H=X$ lies there. Thus [L1] makes $H$ Carathéodory measurable in either case.

One has $E\subseteq H$. In the finite case, monotonicity and each covering bound give ${\mu }^{\ast }(E)\le {\mu }^{\ast }(H)\le {\mu }^{\ast }(U_n)<{\mu }^{\ast }(E)+1\mathrm{/}(n+1)$ for every $n$, hence ${\mu }^{\ast }(H)={\mu }^{\ast }(E)$; in the infinite case both values are $+\mathrm{\infty }$. Therefore [F1] makes $H$ a generated measurable hull, and ${\mu }^{\ast }$ is regular.

For the forward direction, suppose $E$ is Carathéodory measurable. Choose by [F1] a cover $(C_k)$ of $E$ with finite cost below ${\mu }^{\ast }(E)+\epsilon \mathrm{/}4$ and put $H={\bigcup }_k{C}_k$. Applying the Carathéodory identity for $E$ to the test set $H$ shows that ${\mu }^{\ast }(H\setminus E)<\epsilon \mathrm{/}4$; the finite total covering cost also has a tail below $\epsilon \mathrm{/}4$, so some finite union $A={\bigcup }_{k<n}{C}_k\in {\mathcal{A}}_0$ satisfies ${\mu }^{\ast }(H\setminus A)<\epsilon \mathrm{/}4$.

For the forward direction, step 1.1 gives $E\mathrm{△}A\subseteq (H\setminus E)\cup (H\setminus A)$ and hence ${\mu }^{\ast }(E\mathrm{△}A)<\epsilon$. For the reverse direction, assume the approximation property, fix a test set $T$, and choose $A\in {\mathcal{A}}_0$ with ${\mu }^{\ast }(E\mathrm{△}A)<\delta$; since ${\mu }^{\ast }$ is an outer measure by [L3] and $A$ is measurable by [L1], subadditivity gives ${\mu }^{\ast }(T\cap E)+{\mu }^{\ast }(T\setminus E)\le {\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\setminus A)+2\delta ={\mu }^{\ast }(T)+2\delta$.

For the reverse direction, letting $\delta$ decrease to $0$ in step 2.1 gives ${\mu }^{\ast }(T\cap E)+{\mu }^{\ast }(T\setminus E)\le {\mu }^{\ast }(T)$, and [L2] applied to the outer measure of [L3] gives the opposite inequality; thus the Carathéodory identity holds for every $T$, completing both implications.

If $(A_k)$ is any algebra cover of $E$, monotonicity and [L3] give $\nu (E)\le \nu ({\bigcup }_k{A}_k)\le {\sum }_k\nu (A_k)={\sum }_k{\mu }_0(A_k)$. Taking the infimum and using the given identity $\mu ={\mu }^{\ast }{\mathrm{\mid }}_{\sigma ({\mathcal{A}}_0)}$ gives $\nu (E)\le {\mu }^{\ast }(E)=\mu (E)$.

Countable choice makes $\mu$ a measure on $\sigma ({\mathcal{A}}_0)$ extending ${\mu }_0$ by [L0]. Suppose $\mu (E)<+\mathrm{\infty }$ and choose an algebra cover $(A_k)$ of $E$ with finite total cost. For $H={\bigcup }_k{A}_k$ and $H_n={\bigcup }_{k<n}{A}_k$, both extensions agree on every $H_n\in {\mathcal{A}}_0$, so [L1] gives $\mu (H)=\nu (H)$; the finite covering cost makes this common value finite.

Apply [L2] to $E\subseteq H$ for each measure. Step 1.1 applied to $H\setminus E$ gives $\nu (H\setminus E)\le \mu (H\setminus E)$, so finite subtraction from the common value in step 2.1 gives $\mu (E)\le \nu (E)$; combined with step 1.1, this proves equality.

Put $P_n={\bigcup }_{k\le n}{A}_k$. By [F2], $P_n\in {\mathcal{A}}_0$, the sequence is increasing with union $X$, and finite additivity from [F1] gives ${\mu }_0(P_n)\le {\sum }_{k\le n}{\mu }_0(A_k)<+\mathrm{\infty }$.

The algebra ${\mathcal{A}}_0$ is nonempty and closed under intersections by [F2], hence is a pi-system by [F3]; it generates $\sigma ({\mathcal{A}}_0)$, and both extensions agree with ${\mu }_0$ on it and on the exhaustion $(P_n)$ from step 1.1.

Applying [L1] to the generating pi-system and the increasing finite-measure exhaustion of step 2.1 gives $\mu =\nu$ on $\sigma ({\mathcal{A}}_0)$.

Since ${\mu }_0(X)<+\mathrm{\infty }$, the constant sequence $P_n=X$ is an increasing sigma-finite exhaustion, including when $X=\mathrm{\varnothing }$ or ${\mu }_0(X)=0$.

The sigma-finite uniqueness theorem [L1] applied to the exhaustion in step 1.1 shows that the finite premeasure has at most one extension to $\sigma ({\mathcal{A}}_0)$.