The Measure Problem and the Continuum Hypothesis

15 Jan 2026

This post was originally written in Korean, and has been machine translated into English. It may contain minor errors or unnatural expressions. Proofreading will be done in the near future.

Definition. When a measure $\mu$ is defined for all subsets of a set $X$, $\mu$ is called a measure on $X$.

Theorem. For any $A \subseteq \mathbb{R}$ and $x \in \mathbb{R}$, there does not exist a measure $\mu$ on the real numbers satisfying $\mu(A + x) = \mu(A)$. (Here, $A + x = \lbrace a + x : a \in A\rbrace$.)

Proof. Refer to the construction of the Vitali set.

Thus, a translation-invariant measure on the real numbers does not exist. Instead, one may weaken the translation-invariance condition and ask whether there exists a measure $\mu$ on the real numbers such that $\mu(A) = m(A)$ for Lebesgue measurable sets $A$, where $m$ denotes the Lebesgue measure. This is called the measure problem.

Measure Problem. Does there exist a measure $\mu$ on the real numbers such that $\mu(A) = m(A)$ for any Lebesgue measurable set $A \subseteq \mathbb{R}$?

In fact, the above problem is known to be equivalent to the following generalised problem:

Generalised Measure Problem. Does there exist a non-trivial probabilistic measure $\mu$ on $2^{\aleph_0}$? (A measure is probabilistic if the measure of the whole space is 1.)

If such a measure exists, it is known that the Lebesgue measure can be extended to $\mathcal{P}(\mathbb{R})$ using this measure and a bijection between $\mathbb{R}$ and $2^{\aleph_0}$.

Remark 1. Since the translation-invariance condition has been removed, the existence of a measure on a set $S$ depends solely on the cardinality of $S$. Therefore, in the subsequent discussion, we focus only on measures on (uncountable) cardinals.

Remark 2. It is known that if $\mu$ is a probabilistic measure on $S$, $C = \lbrace x \in S : \mu(\lbrace x \rbrace ) > 0 \rbrace $ is at most countable. Hence if $|S| = \kappa$ is uncountable, we have $|S| = |S \setminus C|$, and $\mu$ restricted to $S \setminus C$ is a measure satisfying $\mu(\lbrace x \rbrace ) = 0$ for all $x \in S \setminus C$. It follows that if a measure $\mu$ exists on an uncountable cardinal $\kappa$, we may assume without loss of generality $\mu(\lbrace \alpha \rbrace ) = 0$ for all $\alpha < \kappa$. Indeed, we will assume this fact from now on.

Interestingly, the measure problem is related to the continuum hypothesis and large cardinal axioms. Let us first examine the following theorem:

Theorem. If a probabilistic measure exists on $\nu$, then some $\kappa \leq \nu$ is weakly inaccessible.

Proof. Suppose a probabilistic measure $\mu$ exists on $\nu$. The collection $\mathcal{I}$ of null sets with respect to $\mu$ forms an ideal. In particular, it can be easily shown that $\mathcal{I}$ satisfies the following properties:

1. Contains singletons: For any $\xi < \nu$, $\lbrace \xi \rbrace \in \mathcal{I}$.
2. Countably closed: If $\lbrace A_n \rbrace_{n < \omega}$ is a collection of sets in $\mathcal{I}$, then $\bigcup_{n < \omega} A_n \in \mathcal{I}$.
3. Satisfies the countable chain condition: If $\lbrace B_i \rbrace_{i \in J}$ is a collection of pairwise disjoint sets not in $\mathcal{I}$, then $|J| \leq \aleph_0$.

Let $\kappa \leq \nu$ be the smallest cardinal such that there exists an ideal $\mathcal{J}$ on $\kappa$ satisfying the above three conditions. We now prove the following lemma:

Definition. Let $\mathcal{I}$ be an ideal. If for any $\lambda < \kappa$, $\lbrace A_\xi \rbrace_{\xi < \lambda}$ is a collection of sets in $\mathcal{I}$, and $\bigcup A_\xi \in \mathcal{I}$, then $\mathcal{I}$ is said to be $\kappa$-complete.

Lemma. $\mathcal{J}$ is $\kappa$-complete.

Proof. Suppose $\mathcal{J}$ is not $\kappa$-complete. Then there exist some $\lambda < \kappa$ and a collection $\lbrace A_\xi \rbrace_{\xi < \lambda}$ of sets in $\mathcal{J}$ such that $\bigcup A_\xi \notin \mathcal{J}$. Without loss of generality, we may redefine $\lbrace A_\xi \rbrace$ as follows to assume that the sets are pairwise disjoint:

\[A_\xi' = A_\xi - \bigcup_{\eta < \xi} A_\eta\]

(Since $A_\xi’ \subseteq A_\xi$, it follows that $A_\xi’ \in \mathcal{J}$ from $A_\xi \in \mathcal{J}$. Additionally, $A_\xi’$ may be empty, but this does not affect the proof as empty sets are trivially disjoint.)

Define an ideal $\mathcal{K}$ on $\lambda$ as follows:

\[S \in \mathcal{K} \iff \bigcup_{\xi \in S} A_\xi \in \mathcal{J}\]

We verify that $\mathcal{K}$ satisfies the three conditions:

1. Contains singletons: This is immediate from the definition of $\mathcal{K}$.
2. Countably closed: Since $\mathcal{J}$ is countably closed, so is $\mathcal{K}$.
3. Satisfies the countable chain condition: Suppose $\lbrace S_i \rbrace_{i \in J}$ is a collection of pairwise disjoint sets not in $\mathcal{K}$. For each $i \in J$, let $B_i = \bigcup_{\xi \in S_i} A_\xi$. Then $B_i \notin \mathcal{J}$, and $B_i \cap B_j = \varnothing$ for $i \neq j$ (using the disjointness of $\lbrace A_\xi \rbrace$). Thus, $\lbrace B_i \rbrace_{i \in J}$ forms a pairwise disjoint antichain in $\mathcal{J}$, implying that $J$ is countable.

This contradicts the minimality of $\kappa$. Hence, $\mathcal{J}$ is $\kappa$-complete. □

From the above lemma, it follows that $\kappa$ is weakly inaccessible, i.e., an uncountable regular limit cardinal.

$\kappa$ is uncountable.

The singleton containment and countable closure properties imply that if $\kappa$ were countable, $\kappa \in \mathcal{J}$, leading to a contradiction.

$\kappa$ is regular.

If $\kappa$ were not regular, there would exist some $\lambda < \kappa$ and an increasing sequence $\lbrace \nu_\xi \rbrace_{\xi < \lambda}$ of cardinals less than $\kappa$ such that $\bigcup_{\xi < \lambda} \nu_\xi = \kappa$. By the singleton containment and $\kappa$-completeness properties, each $\nu_\xi$ belongs to $\mathcal{J}$. Hence, by $\kappa$-completeness, $\bigcup_{\xi < \lambda} \nu_\xi = \kappa$ belongs to $\mathcal{J}$, contradicting the fact that $\mathcal{J}$ is an ideal.

$\kappa$ is a limit cardinal.

If $\kappa$ were not a limit cardinal, there would exist some ordinal $\alpha$ such that $\kappa = \aleph_{\alpha + 1}$. For each $\gamma < \kappa$, there exists a surjection $f_\gamma: \omega_\alpha \to \gamma$. Using the axiom of choice, we obtain a sequence of such functions $\lbrace f_\gamma \rbrace_{\gamma < \omega_{\alpha + 1}}$.

For each $\beta < \omega_\alpha$ and $\gamma < \omega_{\alpha + 1}$, define:

\[A_{\gamma\beta} = \{ \xi < \omega_{\alpha + 1} : f_\xi(\beta) = \gamma \}\]

Observe the following:

• $\bigcup_{\beta < \omega_\alpha} A_{\gamma\beta} = \omega_{\alpha + 1} - (\gamma + 1)$
• $\gamma_1 \neq \gamma_2 \implies A_{\gamma_1 \beta} \cap A_{\gamma_2 \beta} = \varnothing$

In particular, by $\kappa$-completeness, $\gamma + 1 \in \mathcal{J}$ implies $\bigcup_{\beta < \omega_\alpha} A_{\gamma \beta} \notin \mathcal{J}$. Hence, for some $\beta$, $A_{\gamma\beta} \notin \mathcal{J}$. Choosing such $\beta_\gamma$ for each $\gamma$, we obtain $\lbrace A_{\gamma\beta_\gamma} \rbrace_{\gamma < \omega_{\alpha + 1}}$. However, the possible values of $\beta_\gamma$ are limited to $\omega_\alpha$, so by the pigeonhole principle, there exists some $\beta < \omega_\alpha$ such that $|\lbrace A_{\gamma\beta_\gamma} : \beta_\gamma = \beta \rbrace| = \aleph_{\alpha + 1}$. This contradicts the countable chain condition. ■

Corollary. If the measure problem has a positive solution, then the continuum hypothesis has a negative solution.

Proof. If the measure problem holds and $2^{\aleph_0} = \aleph_1$, then by the theorem, $\aleph_0$ or $\aleph_1$ must be weakly inaccessible. However, $\aleph_0$ is countable, and $\aleph_1$ is a successor cardinal, leading to a contradiction. ■