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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