Measure on Hilbert Space

As clarified, you are looking for translation-invariant Borel measures. Here are two: the zero measure, and counting measure. Obviously those are not going to satisfy you, but you can't really do better.

Theorem. A translation-invariant Borel measure on an infinite-dimensional separable Banach space is either the zero measure, or assigns infinite measure to every open set.

You can find a proof on Wikipedia, or in Theorem 1.1 of these notes I wrote.

The following facts are valid:

Fact 1. There exists a sigma-finite Borel measure in $\ell_2$ which is invariant under the group of all eventually zero sequences and takes the value $1$ on the Hilbert cube $\prod_{k \in N}[0,\frac{1}{k}]$.

The proof of Fact 1 can be found in [Kharazishvili A.B., On invariant measures in the Hilbert space.Bull. Acad. Sci.Georgian SSR, 114(1) (1984),41--48 (in Russian)].

Fact 2. There exists a translation-invariant Borel measure $\mu$ in $\ell_2$ which takes the value $1$ on the Hilbert cube $\prod_{k \in N}[0,\frac{1}{k}]$.

The proof of Fact 2 can be obtained by Baker measure $\lambda$ [Baker R., ``Lebesgue measure" on $\mathbb{R}^{ \infty}$. II. \textit{Proc. Amer. Math. Soc.} vol. 132, no. 9, 2003, pp. 2577--2591] as follows:

Let $T:\ell_2 \to \mathbb{R}^{ \infty}$ be defined by $T((x_k)_{k \in N})=(k x_k)_{k \in N}$ for $(x_k)_{k \in N} \in \ell_2$. For each Borel subset $X \subseteq \ell_2$ we set $\mu(X)=\lambda(T(X))$. Then $\mu$ satisfies all conditions of Fact 2.

Some authors take local compactness of the space to be part of the definition of Borel measure, so that leaves out infinite-dimensional Hilbert spaces right away. I think the Mackey-Weil result is talking about $\sigma$-finite measures. If you don't require that, you might consider $r$-dimensional Hausdorff measure for any nonnegative real $r$. These are translation-invariant measures, and all Borel sets are measurable.