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 $\mu = \lambda$.



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


