I was reading a paper and they claim the following. Suppose $$ \int_{\mathbb{R}} g(\lambda x + \theta) d\mu(x) = 0, $$ where $\mu$ is a signed Borel measure, $g$ is continuous and in $S'(\mathbb{R})$. Here $S(\mathbb{R})$ is the Schwartz space, and $S'(\mathbb{R})$ is its dual. Let $w(\theta) \in S(\mathbb{R})$. Then, $$ (1) \quad \int_{\mathbb{R}} w(\theta)d\theta \int_{\mathbb{R}} g(\lambda x + \theta) d\mu(x) = 0, $$ that I interprete it as $$ \int_{\mathbb{R}} \left[w(\theta) \int_{\mathbb{R}} g(\lambda x + \theta) d\mu(x)\right] d\theta = 0. $$ By setting $u = \lambda x + \theta$, and change order of integration, the paper claim that the above is $$ (2) \quad \int_{\mathbb{R}} g(u) \int_{\mathbb{R}} w(\theta) d\mu(\frac{u-\theta}{\lambda}) = 0, $$ that I am not sure what is means. It is claimed that the above is equivalent to $$ (3) \quad \hat{g}(\hat{w}(\cdot)\widehat{d\mu}(\lambda \cdot)) = 0, $$ where $\hat{g}$ is the Fourier transform of $g$ in the sense of tempered distribution.
I am not sure how (2) and (3) are equivalent and what these mean.
Any comments/suggestions/answers will be very appreciated.