Let $(\mathcal{X},d)$ be a metric space and $\mu$ be a Borel probability measure on $(\mathcal{X},d)$. Let $m \in \mathbb{N}$ and define the two probability product measures $\mu^{m} := \otimes_{k=1}^{m}\mu$ and $\mu^{2m} := \otimes_{k=1}^{2m}\mu$ on the Borel subsets of the product spaces $\mathcal{X}^{m}$ and $\mathcal{X}^{2m}$, respectively. Let $\mathcal{C}$ be an arbitrary family of Borel subsets of $(\mathcal{X},d)$ (in general not denumerable, but that we may assume to have finite Vapnik-Chervonenkis dimension) and $\varepsilon>0$. Define $$Q:=\Big\{(x_1,\dots,x_m) \in \mathcal{X}^{m}\ \Big| \ \exists C \in \mathcal{C}, \Big| \mu[C] - \frac{1}{m} \sum_{k=1}^m \chi_C(x_k) \Big| \ge \varepsilon \Big\}.$$
The $\varepsilon$-sample theorem proof has as a first step a symmetrization trick. I have a question about the rigorousness of that part of the proof. It is based on the following step that left me baffled: \begin{align*} \mu^{2m}\bigg[ \Big\{(x_1,\dots,x_m,x_1',\dots,x'_m) \in \mathcal{X}^{2m}\ \Big| \ \exists C \in \mathcal{C}, \Big| \mu[C] - \frac{1}{m} \sum_{k=1}^m \chi_C(x_k) \Big| \ge \varepsilon \land \Big| \mu[C] - \frac{1}{m} \sum_{k=1}^m \chi_C(x_k') \Big| \le \frac{\varepsilon}{2}\Big\} \bigg]\\ =\int_{Q} \mu^m \bigg[ \Big\{ (x_1',\dots,x_m') \in \mathcal{X}^m \ \Big| \ \exists C \in \mathcal{C}, \Big| \mu[C] - \frac{1}{m} \sum_{k=1}^m \chi_C(x_k) \Big| \ge \varepsilon \land \Big| \mu[C] - \frac{1}{m} \sum_{k=1}^m \chi_C(x_k') \Big| \le \frac{\varepsilon}{2} \Big\}\bigg]\operatorname{d}\mu^m(x_1,\dots,x_m) \end{align*}
Here, it is clear that the author of the proof wanted to use disintegration via independence. However, the involved sets are not measurable in general: it is not assumed that the family $\mathcal{C}$ is denumerable, so we have no guarantee about the measurability of every set defined through a condition of the form "$\exists C \in \mathcal{C}$". Then, I thought that the above equation could be interpreted in terms of the corresponding (uniquely defined) outer measures. But, at that point, I'm not even sure that what it is written actually makes sense (we are integrating on a set $Q$ that, in general, is not measurable, and it isn't clear that the integrand function is measurable at all), nevermind if disintegration could be performed anymore.
Is it possible to make the previous step rigorous without assuming the measurability of the involved sets?
That line of the proof, with just slightly different notations, could be found in e.g., Anthony, Bartlett - Neural Network Learning, Theoretical Foundations, Lemma 4.4, pag. 46 and 47, eq. (4.3).