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

Question

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 $\mu = \lambda$.

Answer 1

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.

Answer 2

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 $(1)$.