Let $\mathfrak{L}(\mathbb{R})$ be the collection of Lebesgue measurable sets and $\mathfrak{B}(\mathbb{R})$ be the Borel sets.

Question: Is there a nontrivial signed measure on $\mathfrak{L}(\mathbb{R})$ that is trivial on $\mathfrak{B}(\mathbb{R})$?

Obviously, any positive measure that is trivial on $\mathfrak{B}(\mathbb{R})$ is also trivial on $\mathfrak{L}(\mathbb{R})$, since any Lebesgue measurable set is a subset of a Borel set.

For the signed case, I have tried doing Jordan decomposition but it doesn't seem work. It is hard (if ever possible) to show $(\mu|_{\mathfrak{B}(\mathbb{R})})^+ = \mu^+|_{\mathfrak{B}(\mathbb{R})}$ and $(\mu|_{\mathfrak{B}(\mathbb{R})})^- = \mu^-|_{\mathfrak{B}(\mathbb{R})}$.

In fact, If I can deal this problem by decomposition, there must be something special about Borel sets, since the above equalities do not hold in general. Let $\mathfrak{C} = \{\varnothing,\{0\},\{1\},\{0,1\}\}$, $\mathfrak{D} = \{\varnothing, \{0,1\}\}$. The signed measure $\lambda$ on $\mathfrak{C}$ is defined that $\lambda(\{0\})=1$ and $\lambda(\{1\})=-1$. Then $\lambda|_\mathfrak{D}$ is trivial and the equalities fail.

Background: I am trying to prove (or disprove) that if $\mu$ and $\lambda$ are signed measures on $\mathfrak{L}(\mathbb{R})$, then $\mu|_{\mathfrak{B}(\mathbb{R})} = \lambda|_{\mathfrak{B}(\mathbb{R})}$ implies $\mu = \lambda$.

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    $\begingroup$ Use the Hahn decomposition instead. $\endgroup$ Oct 20, 2020 at 7:21
  • $\begingroup$ If $(H^+,H^-)$ is a Hahn decompostion of $\mu$. Do you mean I should apply the positive case to $\mu(E\cap H^+)$ and $\mu(E\cap H^-)$? $\endgroup$ Oct 20, 2020 at 12:59
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    $\begingroup$ So every Borel subset of $H^+$ or $H^-$ is null... but I don't see how that helps. @MichaelGreinecker can you give another hint? $\endgroup$ Oct 20, 2020 at 13:53
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    $\begingroup$ I guess one note is the following. Suppose a nontrivial measure $\mu$ exists, so $\mu(H^+) > 0$. We can write $H^+ = B^+ \cup N^+$ where $B^+$ is Borel and $N^+$ is null. Since $\mu(B^+) = 0$ by assumption, $\mu(N^+) > 0$. In particular, $N^+$ is not Borel and necessarily uncountable. But every subset of $N^+$ is Lebesgue measurable, so $\mu|_{N^+}$ is a nontrivial countably additive positive measure on $2^{N^+}$. I think this makes $|N^+|$ a real-valued measurable cardinal, or at least implies the existence of one, and this is unprovable in ZFC. $\endgroup$ Oct 20, 2020 at 13:58
  • $\begingroup$ Does @NateEldredge's comment work in reverse? If there is a real-valued measurable cardinal, can we use it to construct an example as in the OP? $\endgroup$ Oct 20, 2020 at 15:45

2 Answers 2


So, promoting my answer to a comment, this is unprovable in ZFC (assuming ZFC is consistent). I claim that such a signed measure $\nu$ exists only if there exists a nontrivial, atomless, countably additive probability measure $\mu$ on the discrete $\sigma$-algebra of $\mathbb{R}$ (or equivalently $[0,1]$). As I understand it, the latter is equivalent to the existence of a real-valued measurable cardinal of size at most $\mathfrak{c}$, which is independent of ZFC.

Suppose such $\nu$ exists. Consider its Hahn decomposition $\mathbb{R} = H^+ \cup H^-$. Since $H^+ \in \mathfrak{L}(\mathbb{R})$, it can be written $H^+ = B^+ \cup N^+$ where $B^+$ is Borel and $N^+$ is Lebesgue-null. By assumption $\nu(B^+) = 0$ so we must have $\nu(N^+) > 0$, and $\nu$ is positive on $N^+$. Now every subset of $N^+$ is Lebesgue measurable, so $\nu$ is defined for every such subset. Thus define $\mu(A) = \nu(A \cap N^+)$ for any subset $A \subset \mathbb{R}$. This is a nontrivial, countably additive, finite, positive measure on $2^{\mathbb{R}}$, which we may rescale to a probability measure. And since singletons are Borel, and therefore have $\nu$-measure zero, $\mu$ is atomless.

Gerald's answer, with Michael's comments, seems to be establishing the converse, that the existence of a real-valued measurable cardinal implies the existence of a desired $\nu$. Combining these would show that the original statement is independent of ZFC.

  • $\begingroup$ Thanks. I see it is impossible to show the existence of such measure in ZFC if ZFC is consistent. However, why is the signed measure $\nu$ constructed in the converse nontrivial on Lebesgue measurable sets? $\endgroup$ Oct 22, 2020 at 13:18
  • $\begingroup$ @Zhang: Responded on the other answer. $\endgroup$ Oct 22, 2020 at 14:10
  • $\begingroup$ Thank you. I have added a new comment below your response. $\endgroup$ Oct 22, 2020 at 14:50

a converse of Nate Eldridge's comment
not a proof, too long for a comment

Suppose there is a real-valued measurable cardinal. We want to show there is a measure as requested.

There is a probability measure $\mu : \mathfrak P([0,1]) \to [0,1]$. We may assume $\mu([0,t]) = t$ for $0 \le t \le 1$.

Using AC of course, can we show the existence of a set $X \subseteq [0,1]$ with $$ \mu(X \cap [0,t]) = t/2\quad \text{for all }t \in [0,1]\quad? \tag1 $$ We can deduce from this: $$ \mu\big(X \cap B\big) = \frac{1}{2}\lambda\big(B\cap[0,1]\big) \quad\text{for all Borel sets }B. \tag2$$
Then the signed measure we want will be $$ \nu(E) = \mu\big(X \cap E\big) - \mu\big((\,[0,1]\setminus X)\cap E\big) $$ From $(2)$ we can prove that $\nu(B) = 0$ for all Borel sets $B$.

Addendum. If we cannot prove $(1)$ for an arbitrary measure $\mu$ as described, maybe we can construct $\mu$ together with $X$ in order to get $(1)$.

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    $\begingroup$ (1) is indeed correct. By a theorem of Gitik-Shelah (found in 5II of Fremlin) the Maharam representation of the measure algebra of $\mu$ is a countable convex combination of uncountable many coin flips each. The Borel $\sigma$-algebra corresponds to countably many factors. In particular, there must be a coin flip independent of all of them. $\endgroup$ Oct 20, 2020 at 22:33
  • $\begingroup$ Your argument works even without the existence of a real-valued measurable cardinal, the non-separable extension of Lebesgue measure constructed by Kakutani and Oxtoby here satisfies (1). $\endgroup$ Oct 20, 2020 at 22:45
  • $\begingroup$ Do you mean $\mu\big(X \cap B\big) = \frac{1}{2}\mu\big(B\cap[0,1]\big)$ in (2)? Also, this might be a silly question, but how can we show that $\nu$ is nontrivial on Lebesgue measurable sets? $\endgroup$ Oct 22, 2020 at 3:34
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    $\begingroup$ @Zhang: I think to get your desired measure, we should push forward the measure $\nu$ from $[0,1]$ to the Cantor set $C$ under a Borel isomorphism $\phi$. The new measure $\phi_* \nu$ is then defined on all subsets of $C$ and zero on all Borel sets, so it must be nonzero on some other subset of the Cantor set, which in particular is Lebesgue measurable. Now you can extend $\phi_* \nu$ to the rest of $\mathbb{R}$ by making it zero outside $C$. $\endgroup$ Oct 22, 2020 at 14:08
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    $\begingroup$ If I correctly understand @MichaelGreinecker's comment, it says that even without a real-valued measurable, there exists a $\sigma$-algebra $\mathcal{F} \supsetneq \mathcal{B}$ and a measure $\mu$ on $\mathcal{F}$, such that there is a set $X \in \mathcal{F}$ satisfying (1). But in order to produce Zhang's desired measure, we would actually need to have $\mathcal{F} = 2^{[0,1]}$ or some similar condition, and it is this which can only happen if there is a real-valued measurable cardinal. $\endgroup$ Oct 22, 2020 at 15:16

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