Change of order of integration with the inner integral being improper and uniform convergent on the integration segment of the outer integral

Question

Suppose $f(x),\phi(x)$ are absolutely integrable on $(-\infty, \infty)$ and $g(\lambda)$ is the Fourier transform of $f(x)$. So we have that $f(x) = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}{g(\lambda)e^{-i\lambda x}d\lambda}$. Let's multiply by $\phi(x)$ and integrate on $[-A,A]$ both sides. According to my book thanks to the uniform convergence on $[-A,A]$ of $\int_{-\infty}^{\infty}{g(\lambda)e^{-i\lambda x}d\lambda}$ we can change integration order and get $\int_{-\infty}^{\infty} {f(x)g(x)dx} = \dfrac{1}{2\pi} \int_{-\infty}^{\infty}{\overline{\big[\int_{-A}^{A}{\phi(x)e^{-i \lambda x}dx}\big ]}g(\lambda)d\lambda}$ where the overline means complex conjugate. I don't understand though how the book reaches such a conclusion. I guess I'm missing some important property on change of order of infinite integrals. Any hint?