Let $\mu$ be a Radon measure on $\mathbb R^d$ with finite total mass: I guess that it is a tempered distribution on $\mathbb R^d$ and thus one may consider its Fourier transform. Now I guess that its Fourier transform $\hat µ$ is a bounded uniformly continuous function. Are these guesses correct?
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Yes, the Fourier transform of a finite measure is the uniformly continuous bounded function $$ \hat{\mu}(x)= \int_{\mathbb R^d} e^{-i x \cdot y} d \mu(y) =: f(x) $$
Now, if we treat $\mu$ as a tempered distribution, its Fourier transform as a distribution is the regular distribution given by $f$, that is $$ \langle \mathcal{F}(\mu), g \rangle =\int_{\mathbb R^d} f(x) g(x) dx \qquad \forall g \in \mathcal{S}(\mathbb R^d) $$
For distributions with compact support, this is Theorem 12.3 in these notes. Now if you want to extend this to finite measures, you can use the fact that for each $\epsilon >0$ there exists some $A>0$ such that the total mass of $\mu$ outside $[-A,A]^d$ is less than $\epsilon$.
