Let us consider the Minkowski space $(\mathbb{R}^{4},\eta)$ and the mass shell $H_{m}$, $m\ge 0$, given by: \begin{eqnarray} H_{m}:=\{x=(x_{0},x_{1},x_{2},x_{3}) \in \mathbb{R}^{4}: \hspace{0.1cm} x\cdot \tilde{x} = m^{2}, x_{0}>0\} \tag{1}\label{1} \end{eqnarray} where $\tilde{x}$ is defined by $x=(x_{0},x_{1},x_{2},x_{3}) \mapsto \tilde{x} := (x_{0},-x_{1},-x_{2},-x_{3})$ and $x\cdot \tilde{x}$ is the usual Lorentz inner product: $$x \cdot \tilde{x} = \eta(x,x).$$ Further, define $j_{m}: \mathbb{R}^{4}\to \mathbb{R}^{3}$ (if $m=0$, take $\mathbb{R}^{3}\setminus \{0\}$ instead) as the homeomorphism $x=(x_{0},x_{1},x_{2},x_{3}) \mapsto (x_{1},x_{2},x_{3})$. The set function: \begin{eqnarray} \mu_{m}(E) := \int_{j_{m}(E)} \frac{d^{3}x}{\sqrt{m^{2}+|x|^{2}}} \tag{2}\label{2} \end{eqnarray} defines a measure on $H_{m}$.

In the literature, $\mu_{m}$ is called Lorentz invariant measure because it is said to be invariant under Lorentz transformations on $\mathcal{L}_{+}^{\uparrow}$.

I have some really basic questions about this settings, but which are not explicitly stated anywhere I know.

**Q1:** The "Lorentz inner product" $x\cdot\tilde{x}$ might be misleading since $x\cdot \tilde{x}$ is not positive-definite, so it does not define an inner product in mathematical terms. Does it play any role in defining the topology on $\mathbb{R}^{4}$, or is $\mathbb{R}^{4}$ equipped with its usual Euclidean topology? Note that the topology is important because $j_{m}$ is assumed to be an homeomorphism and, also, because $\mu_{m}$ is a measure on $H_{m}$ which, I assume, becomes a measure space when equipped with the Borel $\sigma$-algebra inherited from $\mathbb{R}^{4}$.

**Q2:** I know that $H_{m}$ is invariant under $\mathcal{L}_{+}^{\uparrow}$ because, if $\Lambda \in \mathcal{L}_{+}^{\uparrow}$ we have: (a) $(\Lambda x)_{0} > 0$ whenever $x_{0}>0$, and (b) $\Lambda x \cdot \tilde{\Lambda x} = x\cdot \tilde{x} = m^{2}$, since it is a Lorentz transformation. But what is the precise meaning of $\mu_{m}$ being Lorentz invariant? Is it as usually stated in ergodic theory, i.e. $\mu_{m}(\Lambda^{-1}E) = \mu_{m}(E)$ for every measurable $E$?