I need this as lemma.
Given the Borel space $\mathcal{B}(\mathbb{R})$.
Consider a complex measure: $$\mu:\mathcal{B}(\mathbb{R})\to\mathbb{C}$$
Then one has: $$\int_{-\infty}^{+\infty}e^{it\lambda}\mathrm{d}\mu(\lambda)=0\implies\mu=0$$
How can I prove this?
Let $f\in C^2(\mathbb {R})$ have compact support. Integrating by parts twice shows $\hat f (x) = O(1/x^2)$ at $\infty;$ hence $\hat f \in L^1.$ By the Fourier inversion formula, $f$ is the inverse Fourier transform of $\hat f.$ Thus
$$\int_{\mathbb {R}} f(x)\,d\mu(x) = \int_{\mathbb {R}}\int_{\mathbb {R}}\hat {f}(t)e^{ixt}\,dt \,d\mu(x) = \int_{\mathbb {R}}\hat {f}(t)\int_{\mathbb {R}}e^{ixt} d\mu(x)\,dt = 0.$$
Because the set of such $f$'s is dense in $C_0(\mathbb {R)},$ and the space of finite Borel measures form the dual space of $C_0,$ we see $\mu = 0.$