# Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets

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 $$\mu = \lambda$$.

• Use the Hahn decomposition instead. Commented Oct 20, 2020 at 7:21
• If $(H^+,H^-)$ is a Hahn decompostion of $\mu$. Do you mean I should apply the positive case to $\mu(E\cap H^+)$ and $\mu(E\cap H^-)$? Commented Oct 20, 2020 at 12:59
• So every Borel subset of $H^+$ or $H^-$ is null... but I don't see how that helps. @MichaelGreinecker can you give another hint? Commented Oct 20, 2020 at 13:53
• 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. Commented Oct 20, 2020 at 13:58
• 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? Commented Oct 20, 2020 at 15:45

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.

• 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? Commented Oct 22, 2020 at 13:18
• @Zhang: Responded on the other answer. Commented Oct 22, 2020 at 14:10
• Thank you. I have added a new comment below your response. Commented 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 $$(1)$$.

• (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. Commented Oct 20, 2020 at 22:33
• 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). Commented Oct 20, 2020 at 22:45
• 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? Commented Oct 22, 2020 at 3:34
• @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 $C$. Commented Oct 22, 2020 at 14:08
• 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. Commented Oct 22, 2020 at 15:16