Bounds on discrepancy metric of product measures Consider two measurable spaces $X_1 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu_1)$ and $X_2 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu_2)$ and the product spaces
$$X_1^{q} = (\times_{i=1}^q\mathbb{R}^m,\otimes_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu_1^{\otimes q})\ \ \ \text{ and }\ \ \ X_2^{q} = (\times_{i=1}^q\mathbb{R}^m,\otimes_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu_2^{\otimes q})$$
where
$\mu_1^{\otimes q} = \underbrace{\mu_1\otimes \cdots \otimes\mu_1}_{q}$ and $\mu_2^{\otimes q} = \underbrace{\mu_2\otimes \cdots \otimes\mu_2}_{q}$, respectively.
Let
$$
D(\mu,\nu) = \sup_{\substack{\text{all closed balls } B \text{ of }\mathbb{R}^m\\ \text{in the Euclidean norm} }} \left|\mu(B)-\nu(B)\right|
$$
be the discrepancy metric between probability measures $\mu$ and $\nu$ (see e.g. this paper for more details on this metric).

My question. For probability measures $\mu_1$ and $\mu_2$, does there exist $k>0$ independent of $\mu_1$ and $\mu_2$
such that
$$
D(\mu_1^{\otimes q},\mu_2^{\otimes q})\le k D(\mu_1,\mu_2)\ \ \ ?
$$

Note. For other probability metrics the answer is in the affirmative (e.g., for the total variation metric, see Eq. (4.5) of this paper) but I couldn't find anything about the discrepancy metric above. I suspect that, if true, this should be a rather known result. Any suggestion or comment is very welcome.
 A: Analogous to the TV metric, the requested upper bound holds for the discrepancy metric with $k=q$.  The result given below can also be easily extended to general product probability measures $\mu=\otimes_{i=1}^q \mu_{i}$ and $\nu=\otimes_{i=1}^q \nu_i$ to obtain $$
D(\mu,\nu)\le \sum_{i=1}^q D(\mu_i, \nu_i)\;.
$$

Suppose that $\mu_1$ and $\mu_2$ are probability measures. Let $B = \{  (x_1, \dots, x_q) : \sum_i (x_i - c_i)^2 \le R^2 \}$ be a closed ball in $\mathbb{R}^q$ with radius $R>0$ centered at $(c_1, \dots, c_q) \in \mathbb{R}^q$.  Let $\mathbf{1}_A$ denote the indicator function of the set $A$.
By telescoping, and invoking Tonelli’s theorem to write $(\mu_1^{\otimes q}-\mu_2^{\otimes q})(B)$ as an iterated integral, note that
\begin{align*} & |(\mu_1^{\otimes q}-\mu_2^{\otimes q})(B)|=
 \left| \int_{\mathbb{R}^q} \mathbf{1}_B(x_1, \dots, x_q) \biggl( d\mu_1(x_1) \cdots d\mu_1(x_q) - d\mu_2(x_1) \cdots d\mu_2(x_q) \biggr) \right|  \\
  &=  \left| \int_{\mathbb{R}^q}  \mathbf{1}_B(x_1, \dots, x_q) \sum_{ i=1}^q d\mu_1(x_1) \cdots d\mu_1(x_{i-1}) \biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) d\mu_2(x_{i+1}) \dots d\mu_2(x_q) \right|  \\ 
  &=  \left| \sum_{ i=1}^q \int_{\mathbb{R}^{q-1}}   \left(\int_{\mathbb{R}} \mathbf{1}_B(x_1, \dots, x_q)\biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) \right) d\mu_1(x_1) \cdots d\mu_1(x_{i-1})  d\mu_2(x_{i+1}) \dots d\mu_2(x_q) \right|  \\ 
&\le   \sum_{ i=1}^q \int_{\mathbb{R}^{q-1}} \left| \int_{\mathbb{R}} \mathbf{1}_B(x_1, \dots, x_q)\biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) \right| d\mu_1(x_1) \cdots d\mu_1(x_{i-1})  d\mu_2(x_{i+1}) \dots d\mu_2(x_q)  \\ 
&\le \sum_{i=1}^q \sup_{\gamma_i, \rho_i} \left| \int_{\mathbb{R}}\mathbf{1}_{(x_i-\gamma_i)^2 \le \rho_i^2} \biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i}) \biggr) \right|  \\
& \qquad \int_{\mathbb{R}^{q-1}} d\mu_1(x_1) \cdots d\mu_1(x_{i-1})  d\mu_2(x_{i+1}) \cdots d\mu_2(x_q) \\
&\le q D(\mu_1, \mu_2) \;. 
\end{align*}
Since $B$ is arbitrary, the conjectured upper bound holds with $k=q$.

Remark. One can do much better than this linear upper bound in some special cases.  For instance, suppose that $\mu_1 = \mathcal{N}(0,1)$ and $\mu_2 = \mathcal{N}(0, \sigma^2)$.  Then, by moving to  hyperspherical coordinates, it's not too hard to show that $$
D(\mu_1^{\otimes q},\mu_2^{\otimes q}) = \left| \frac{\Gamma(\frac{q}{2}, \frac{q \log{\sigma}}{\sigma^2-1}) - \Gamma(\frac{q}{2}, \frac{q \sigma^2 \log(\sigma)}{\sigma^2-1})}{ \Gamma(\frac{q}{2}) } \right| \; $$ which converges to one with $q$ and $k(q):= D(\mu_1^{\otimes q},\mu_2^{\otimes q}) /D(\mu_1,\mu_2) $ grows sublinearly.  This Gaussian case is rather exceptional because the corresponding product measure can be directly written in terms of the Euclidean distance.
