Let me denote by $x=(x_{0},x_{1},x_{2},x_{3})$ a generic point in $\mathbb{R}^{4}$ and by ${\bf{x}} = (x_{1},x_{2},x_{3})$ the "spatial coordinates" of $x$ in $\mathbb{R}^{3}$, so that $x \in \mathbb{R}^{4}$ becomes $x = (x_{0},{\bf{x}})$. We can define the projection $P: \mathbb{R}^{4} \to \mathbb{R}^{3}$ by setting $P(x_{0},{\bf{x}}) := {\bf{x}}$. Let us define a measure on the Borel sets of $\mathbb{R}^{4}$ by: $$\lambda_{m}(E) := \int_{P(E)}\frac{d{\bf{x}}}{\sqrt{m^{2}+|{\bf{x}}|^{2}}} $$ where $m > 0$ is just a fixed constant. The $d{\bf{x}}$ term means integration with respect to the variables $x_{1},x_{2},x_{3}$. This measure is called Lorentz invariant measure in the physics literature.
Question: I'm trying to understand what is the correct expression to integrate functions with respect to $\lambda_{m}$. More concretely, suppose $f: \mathbb{R}^{4} \to \mathbb{R}$ is measurable and $\lambda_{m}$-integrable. How can we express the integral: $$\int_{\mathbb{R}^{4}}f(x)d\lambda_{m}(x)$$ in terms of the density $d{\bf{x}}/\sqrt{m^{2}+|{\bf{x}}|^{2}}$ of $\lambda_{m}$?
Some thoughts: We can write $\lambda_{m}$ alternatively as: $$\lambda_{m}(E) = \int_{P(E)} \int_{\mathbb{R}}\frac{1}{||x||}\delta(m^{2}-y)d^{4}x$$ where $||x|| = \sqrt{x_{0}^{2}+\cdots + x_{3}^{2}}$ and $d^{4}x$ is the integral with respect to all variables $x_{0},x_{1},x_{2},x_{3}$. Hence, the answer seems to be: $$\int_{\mathbb{R}^{4}}f(x)d\lambda_{m}(x) = \int_{\mathbb{R}^{4}}f(x) \frac{1}{||x||}\delta(m^{2}-y)d^{4}x = \int_{\mathbb{R}^{3}} f(m^{2},{\bf{x}})\frac{1}{\sqrt{m^{2}+|{\bf{x}}|^{2}}}d{\bf{x}}$$
Do my calculations seem correct?
$\lambda_m$ as you have defined it is not even a measure because it's not even additive (let alone countably additive). For example, let $A\subset\Bbb{R}^3$ be a Borel set, and for each $t\in\Bbb{R}$ let $E_t=\{t\}\times A$. This is a "slice of $A$ at height $t$". Clearly, $E_0$ and $E_1$ are disjoint sets, but according to your definition for $\lambda_m$, only the projection to $\Bbb{R}^3$ matters, so you have \begin{align} \lambda_m(E_0\cup E_1)=\lambda_m(E_0)=\lambda_m(E_1) \end{align} and this is NOT equal to $\lambda_m(E_0)+\lambda_m(E_1)$ if $A$ has positive $3$-dimensional Lebesgue measure.
What we ought to do is first define a measure $\zeta_m$ on Borel subsets of $\Bbb{R}^3$ (NOT $\Bbb{R}^4$) by the formula \begin{align} \zeta_m(A)&:=\int_A\frac{d^3\mathbf{x}}{\sqrt{m^2+\|\mathbf{x}\|^2}}. \end{align} Next, we let $\delta_{m^2}$ be the Dirac measure on $\Bbb{R}$, concentrated at $m^2$. Note that $\delta_{m^2}$ is a finite measure, and $\zeta_m$, while it's an infinite measure, it is $\sigma$-finite. Therefore, we consider the product Borel measure on $\Bbb{R}^4$ given by $\lambda_m:=\delta_{m^2}\times \zeta_m$. Now, for any Borel-measurable $f:\Bbb{R}^4\to [0,\infty]$, Tonelli's theorem tells us that \begin{align} \int_{\Bbb{R}^4}f\,d\lambda_m&=\int_{\Bbb{R}^3}\int_{\Bbb{R}}f(y,\mathbf{x})\,d\delta_{m^2}(y)\,d\zeta_m(\mathbf{x})\\ &=\int_{\Bbb{R}^3}f(m^2,\mathbf{x})\,d\zeta_m(\mathbf{x})\\ &=\int_{\Bbb{R}^3}\frac{f(m^2,\mathbf{x})}{\sqrt{m^2+\|\mathbf{x}\|^2}}\,d^3\mathbf{x}. \end{align}
In fancier terms, what you tried to do is take the measure $\zeta_m$ and use the projection $P$ to get a "pullback measure" $P^*(\zeta_m)$. But in general, measures are pushed-forward. We can only pull-back the measure if the function and its inverse are measurable, in which case pull-back is defined as push-forward by the inverse mapping. In your case, $P$ is not even invertible, so this procedure is bound to fail.