Let $E$ be a separable $\mathbb R$-Banach space and $\lambda\_i$ be a finite symmetric measure on $\mathcal B(E)$ with $\lambda\_i(\{0\})=0$ and $$\int\_B1-\cos\langle x,x'\rangle\:\underbrace{(\lambda\_1-\lambda\_2)}\_{=:\:\sigma}({\rm d}x)=0\tag1$$ for all $B\in\mathcal B(E)$ and $x'\in E'$.



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> How can we conclude that $\lambda\_1=\lambda\_2$?
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The idea is the following: Let $$0\le f\_{x'}(x):=\sum\_{n\in\mathbb N}\frac{1-\cos\frac{\langle x,x'\rangle}n}{2^{n+1}}\le1\;\;\;\text{for }x\in E$$ for $x'\in E'$ and note that $$f\_{x'}(x)=0\Leftrightarrow\langle x,x'\rangle=0.\tag2$$ Let $(x\_n)\_{n\in\mathbb N}\subseteq E\setminus\{0\}$ be dense and $(x'\_n)\_{n\in\mathbb N}\subseteq E'$ with $\|x'\_n\|\_{E'}=1$ and $\langle x\_n,x'\_n\rangle=\left\|x\_n\right\|\_E$ for all $n\in\mathbb N$. Now let $$0\le g(x):=\sum\_{n\in\mathbb N}\frac{f\_{x\_n'}(x)}{2^n}\le1\;\;\;\text{for }x\in E$$ and note that $$g(x)=0\Leftrightarrow x=0\tag3.$$ By Lebesgue's dominated convergence theorem and $(1)$, $$\int\_Bg\:{\rm d}\sigma=0\tag4\;\;\;\text{for all }B\in\mathcal B(E).$$



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> How can we conclude that $\sigma(B)=0$ for all $B\in\mathcal B(E)$ with $0\not\in B$?
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If $\lambda_i$ is symmetric with $\lambda_i\{0\}=0$, why does $\int_B1-\cos\langle x,x'\rangle\:(λ_1-λ_2)({\rm d}x)=0$ imply $λ_1=λ_2$?
$\newcommand\B{\mathcal B}$Let $C:=E\setminus\{0\}$, so that $g>0$ on $C$. Let $\rho(B):=\sigma(B)$ for all $B\in\B(C)$, so that $\rho$ is a signed measure defined on $\B(C)$ such that
$$\int\_B g\,d\rho=0 \tag{1}$$
for all $B\in\B(C)$.


By the [Hahn decomposition theorem](https://en.wikipedia.org/wiki/Hahn_decomposition_theorem), $C$ can be partitioned into two sets, $P$ and $N$ in $\B(C)$ so that $P$ is a positive set for $\rho$ and $N$ is a negative set for $\rho$.
Substituting $P$ for $B$ in (1), we get $\rho(P)=0$. Similarly, $\rho(N)=0$.


So, $\rho=0$; that is, $\sigma(B)=0$ for all $B\in\B(E)$ with $0\notin B$, as desired.


