If the Fourier transform of a measure is zero then the measure is zero 
If $\mu$ is a complex finite Borel measure on a separable real Hilbert space $H$ be such that $$\hat \mu (x) = \int \limits _H \Bbb e ^{\Bbb i \langle x, y \rangle} \Bbb d \mu _{(y)} = 0, \ \forall x \in H ,$$ then $\mu = 0$.

I have found a similar problem on $\Bbb R$ but the hints given there do not help me much in this case. The authors of the book where I have found this suggest the following:
1) show that $\mu \big( \{ y \in H \mid \langle x, y \rangle \ge \alpha \} \big) = 0$
2) it follows that the measure of every closed convex set is $0$
2') in particular, the measure of closed balls is $0$
3) then the measure of every strongly measurable set is $0$
4) so $\mu = 0$.
I do not understand almost anything of the above, my knowledge of measure theory being at an undergraduate level. In particular, I do not know what a strongly measurable set is. Could anyone please at least sketch a proof, following the above lines or an alternate idea? (The context is to show that the Banach algebra of complex finite measures embeds topologically in the Banach algebra of complex continuous bounded functions.)
 A: Assume we already know that the Fourier transform is determining on $\mathbb R$ (i.e. we know how to address the case $H=\mathbb R$).
Let us follow the hints in the OP. In order to show (1), fix $x\in H$ and define $x^*\colon y\to \langle x|y\rangle$. The function $x^*$ is (weakly = strongly) measurable (see below), and thus we may consider the push-forward measure $x^*_\sharp\mu$, a finite complex measure on $\mathbb R$. Thus,
$$\widehat{x^*_\sharp\mu}(s)= \int e^{i s t} dx^*_\sharp\mu(t)=\int e^{i s x^*(y)}d\mu(y)=\hat\mu(sx)=0 \qquad s>0, \; x\in H.$$
Since the Fourier transform is determining on $\mathbb R$, we conclude that $x^*_\sharp \mu=0$ is the $0$-measure. By definition, $\mu \{x^*(y)\geq \alpha\}=x^*_\sharp\mu [\alpha,\infty)=0$.
Every closed convex set is an intersection of half-spaces (consequence of Hahn-Banach, see this question). Since $H$ is separable, $H^*$ is separable, and thus the intersection can be taken to be at most countable.
This shows that (1) implies (2).
Since $\mu$ is a finite measure on a complete separable metric space, it is outer regular, thus (2') directly implies (4).
Side note: on "strongly measurable".
There are several possible meaning of the phrasing "strongly measurable", depending on context. Here, the authors probably mean "measurable w.r.t. the Borel $\sigma$-algebra of the norm (i.e. strong) topology", as opposed to "weakly measurable", i.e. w.r.t. the Borel $\sigma$-algebra of the weak (or weak*, depending on context) topology. In infinite dimensions the distinction is relevant (only) on non-separable Banach spaces, see this question.
Incidentally, googling "strongly measurable set" one finds out that their use is virtually a hapax (there is a homonymous unrelated notion in algebraic geometric though). Also, the same phrasing is used in the "original" version of the book, these notes by S. Albeverio and R. Høegh-Krohn.
A: It seems to me that if $V$ is any finite dimensional subspace, and $V^\perp$ its orthogonal complement, then we can define a measure $\mu_V$ on $V$ by
$$ \mu_V(A) = \mu(A + V^\perp) $$
and we can calculate that $\hat \mu_V = \hat\mu \big|_V $.  So by the finite dimensional result, $\mu_V = 0$ for all finite dimensional V.
Let $V_n$ be an ascending sequence of finite dimensional vector spaces whose union is dense in $H$.  Hence $V_n^\perp$ is a descending sequence of finite codimension spaces whose intersection is $\{0\}$.  Then by the Lebesgue dominated convergence theorem
$$ \mu(A) = \lim_{n\to\infty} \mu(A + V_n^\perp) ,$$
and
$$ \mu(A + V_n^\perp) = \mu_{V_n}((A + V_n^\perp)\cap V_n) = 0 .$$
Hence $\mu = 0$.
