Let $\mu$ be a finite measure on $(\mathbf{R},\mathcal{B})$. Define $\hat\mu(t)=\int{e^{-itx} d\mu(x)} , t\in \mathbf{R}$. Show that the function $\hat\mu$ is uniformly continuous on $\mathbf{R}$. $\hat\mu$ is called fourier transform of $\mu$.
I think I should use Bounded Convergence Theorem. That should hold for Complex valued functions as well. So, we get if $t_n\rightarrow t$ as $n\rightarrow \infty, \int|e^{-it_n x}-e^{-itx}|d\mu(x)\rightarrow 0 $ as $n\rightarrow \infty$ as $| e^{-it_n x}|\leq 1$. Does this imply uniform continuity? I can't see that.