Common Errors in Caratheodory Proofs

Where Caratheodory Proofs Go Wrong

Many errors in proofs about outer measures come from using familiar measure identities before their hypotheses have been established. An outer measure is monotone and countably subadditive, but it need not be additive on disjoint sets. The Caratheodory criterion identifies the sets for which a precise splitting equality does hold. Even then, if outer-measure values can be infinite, the equality cannot be rearranged by ordinary subtraction.

Let $X$ be a set and let $\mu^*:\mathcal{P}(X)\to[0,\infty]$ be an outer measure. This means $\mu^*(\varnothing)=0$, $\mu^*(A)\leq\mu^*(B)$ whenever $A\subseteq B$, and for every sequence $(A_n)$ of subsets of $X$, $\mu^*(\bigcup_n A_n)\leq\sum_n\mu^*(A_n)$. No measure space is assumed here: $\mu^*$ is defined on all subsets of $X$, but is not generally additive.

Definition: A set $E\subseteq X$ is Caratheodory measurable with respect to $\mu^*$ if, for every $A\subseteq X$, $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E). $$ The sets satisfying this criterion form a sigma-algebra, and the restriction of $\mu^*$ to that sigma-algebra is a measure. These are standard conclusions of Caratheodory's theorem, established earlier in the course.

For every $E\subseteq X$, whether measurable or not, countable subadditivity applied to $A=(A\cap E)\cup(A\setminus E)$ gives $\mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\setminus E)$. Thus the substantive part of the measurability test is the reverse inequality. A proof that only obtains the subadditive direction has not established the criterion.

Disjoint Does Not Mean Additive

The first common mistake is to reason that disjointness alone implies additivity. For a general outer measure, disjoint sets need not have additive outer measures. Additivity is available for disjoint sets that are Caratheodory measurable, after the criterion has been verified. Using it to prove that the sets are measurable assumes the point at issue.

Worked Example: A Disjoint Split That Is Not Additive

Take $X=\{a,b\}$, and define $\mu^*(\varnothing)=0$ and $\mu^*(A)=1$ for every nonempty $A\subseteq X$. First check that this is an outer measure. It is monotone, and it vanishes at the empty set. If a union is empty, every set in its cover is empty and subadditivity reads $0\leq0$. If the union is nonempty, at least one set in any cover is nonempty, so the sum of the outer measures in the cover is at least $1$, while the union has outer measure $1$. Thus countable subadditivity holds.

The disjoint sets $\{a\}$ and $\{b\}$ have outer measures $1$ and $1$, but their union has outer measure $1$. In particular, for $E=\{a\}$ and $A=X$, the two sides of the Caratheodory test are $\mu^*(X)=1$ and $\mu^*(X\cap E)+\mu^*(X\setminus E)=1+1=2$. Therefore $E$ is not measurable. Check: the test does hold for $E=\varnothing$ and $E=X$, since the split is respectively $\varnothing\cup A$ and $A\cup\varnothing$, giving $0+\mu^*(A)=\mu^*(A)$ in either case.

A related valid step is finite splitting by sets already known to be measurable. The order matters: use the defining criterion at each step, rather than assuming finite additivity for arbitrary pieces.

Proposition (Finite splitting): Let $E_1,\ldots,E_N$ be pairwise disjoint Caratheodory-measurable subsets of $X$, where $N\geq1$. For every $A\subseteq X$, $$ \mu^*(A)=\sum_{j=1}^{N}\mu^*(A\cap E_j) +\mu^*\!\left(A\setminus\bigcup_{j=1}^{N}E_j\right). $$ All terms and sums are interpreted in $[0,\infty]$.

Proof. Define $B_0=A$ and $B_j=A\setminus\bigcup_{i=1}^{j}E_i$ for $1\leq j\leq N$. Apply the Caratheodory criterion for $E_1$ to $B_0$: $\mu^*(B_0)=\mu^*(B_0\cap E_1)+\mu^*(B_1)$. For each $j$ with $2\leq j\leq N$, apply the criterion for $E_j$ to $B_{j-1}$. Pairwise disjointness gives $B_{j-1}\cap E_j=A\cap E_j$, and $B_{j-1}\setminus E_j=B_j$. Hence $\mu^*(B_{j-1})=\mu^*(A\cap E_j)+\mu^*(B_j)$. Substituting these equalities successively gives the stated formula. This uses only addition of nonnegative extended-real numbers; it does not subtract any quantities. $\square$

Why Infinite Outer Measures Cannot Be Subtracted

A second error is to rearrange a splitting equation as though all its terms were finite real numbers. In $[0,\infty]$, addition is defined, but subtraction of infinite quantities is not: $\infty-\infty$ has no defined value. Even a valid equality with both sides infinite may provide no information about the finite contribution of one part.

Worked Example: A Valid Equality That Cannot Be Rearranged

Use Lebesgue outer measure $m^*$ on $\mathbb{R}$, and split $A=\mathbb{R}$ using $E=\mathbb{Q}$. The rationals are countable, so $m^*(\mathbb{Q})=0$. Also $m^*(\mathbb{R})=\infty$: for each positive integer $N$, monotonicity gives $m^*(\mathbb{R})\geq m^*([-N,N])=2N$, and these lower bounds have no finite upper bound. Since $\mathbb{R}=\mathbb{Q}\cup(\mathbb{R}\setminus\mathbb{Q})$, countable subadditivity implies $\infty=m^*(\mathbb{R})\leq 0+m^*(\mathbb{R}\setminus\mathbb{Q})$; hence $m^*(\mathbb{R}\setminus\mathbb{Q})=\infty$.

The splitting equation is $\infty=0+\infty$, which is correct. But trying to recover the first piece by subtracting the second would require evaluating $\infty-\infty$, which is undefined; it cannot be used to derive $m^*(\mathbb{Q})=0$. That value was obtained from countability, not subtraction. Check: substituting the independently established values into the criterion gives $\infty=0+\infty$, as required for this particular split. This one split does not by itself prove that $\mathbb{Q}$ is measurable; the criterion must hold for every $A\subseteq\mathbb{R}$.

If $\mu^*(A)<\infty$, then monotonicity ensures both pieces of any split of $A$ have finite outer measure. In that case, ordinary subtraction can be used after a valid splitting equality has been established. This does not make arbitrary disjoint sets additive: the measurability condition is still essential.

Subadditivity Gives Only One Inequality

A third error arises when a proof invokes countable subadditivity and then writes an equality. For any sequence $(A_n)$, subadditivity gives only $\mu^*(\bigcup_n A_n)\leq\sum_n\mu^*(A_n)$. To prove equality for disjoint measurable sets, one needs the opposite inequality as well. The finite splitting proposition supplies it: split the union by the first $N$ sets, retain the nonnegative remainder, and then let $N$ increase. No limit-interchange theorem is involved; the infinite sum of nonnegative terms is defined as the limit, equivalently the supremum, of its partial sums.

Theorem (Countable additivity on Caratheodory sets): Let $(E_n)_{n\geq1}$ be pairwise disjoint Caratheodory-measurable subsets of $X$. Then their union $U=\bigcup_{n\geq1}E_n$ is Caratheodory measurable, and $$ \mu^*(U)=\sum_{n=1}^{\infty}\mu^*(E_n). $$ The union's measurability follows from the sigma-algebra property of the Caratheodory-measurable sets. The equality holds in $[0,\infty]$.

Proof. Set $U_N=\bigcup_{n=1}^{N}E_n$. Apply finite splitting to $A=U$ and $E_1,\ldots,E_N$. Since $E_n\subseteq U$ and the sets are disjoint, $U\cap E_n=E_n$. The formula gives $\mu^*(U)=\sum_{n=1}^{N}\mu^*(E_n)+\mu^*(U\setminus U_N)$, so $\mu^*(U)\geq\sum_{n=1}^{N}\mu^*(E_n)$ for every $N$. Taking the supremum over $N$ yields $\mu^*(U)\geq\sum_{n=1}^{\infty}\mu^*(E_n)$, including the case where the partial sums are unbounded. In the other direction, countable subadditivity applied to $U=\bigcup_nE_n$ gives $\mu^*(U)\leq\sum_{n=1}^{\infty}\mu^*(E_n)$. The two inequalities prove equality. If any term is infinite, the lower bound already forces $\mu^*(U)=\infty$, so the argument also covers that case. $\square$

Worked Example: Strict Subadditivity for Countably Many Sets

Let $X=\mathbb{N}$, and define $\mu^*(\varnothing)=0$ and $\mu^*(A)=1$ for every nonempty $A\subseteq\mathbb{N}$. This is an outer measure: for a nonempty union, at least one member of any cover is nonempty, so the sum of the outer measures is at least $1$; the empty-union case gives $0\leq0$.

Take $E_n=\{n\}$. These sets are pairwise disjoint, and $\bigcup_nE_n=\mathbb{N}$. Direct calculation gives $\mu^*(\bigcup_nE_n)=1$, while $\sum_n\mu^*(E_n)=\sum_n1=\infty$. Thus the subadditivity inequality is strict: $1\leq\infty$. This does not contradict the theorem, because the singletons are not Caratheodory measurable. For $E=\{1\}$ and $A=\mathbb{N}$, the criterion would require $1=1+1=2$, which is false. Check: substituting the computed values into subadditivity gives $1\leq\infty$; substituting them into the additivity claim would give the false statement $1=\infty$.

A Reliable Proof Check

When reading or writing a Caratheodory argument, identify the exact source of each equality. A splitting equality comes from measurability of the set used to split; an inequality in the subadditive direction comes from the outer-measure axiom; and countable additivity requires both directions. Before rearranging an equation, check that every quantity being subtracted is finite. These checks prevent circular proofs and keep extended-real arithmetic valid.

Common mistake and correction. The wrong calculation is: “The sets $E_n$ are disjoint, so $\mu^*(\bigcup_nE_n)=\sum_n\mu^*(E_n)$.” Disjointness alone does not imply this, as the countable example shows. The correction is to first establish that each $E_n$ is Caratheodory measurable; then finite splitting gives the lower bound by partial sums, while countable subadditivity gives the upper bound. Similarly, from $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)$, do not isolate a term by subtraction unless the relevant values are finite.

Check Your Understanding

For each question, identify which axiom or criterion justifies the proposed step.

For an arbitrary outer measure and a set $E\subseteq X$, which direction of the Caratheodory splitting inequality follows directly from countable subadditivity?
In the two-point example, calculate both sides of the criterion for $E=\{a\}$ and $A=X$. What does the failed equality show?
Why is it invalid to infer a value for one summand by rearranging $\infty=c+\infty$?
In the countable-additivity proof, which step supplies the lower bound for $\mu^*(\bigcup_nE_n)$, and which step supplies the upper bound?
For the outer measure on $\mathbb{N}$ that assigns value $1$ to every nonempty set, why does the sequence of singleton sets not contradict countable additivity on Caratheodory-measurable sets?