Question. If $\mu_1$,$\mu_2$ measures on the space $(\mathbb{R}^n, \mathcal{B}^n)$, where $\mathcal{B}^n$ is the Borel $\sigma$-algebra, and $$\mu_1\left(\prod_{i=1}^n (x_i, y_i]\right) = \mu_2\left(\prod_{i=1}^n (x_i, y_i]\right) \tag{1},$$ with $x_i,y_i \in \mathbb{R}$, then provide a counterexample to show that $\mu_1 \neq \mu_2$.
So far. There's a lemma that I'm aware of that deals with this where if for (1) we instead have, $\mu_1(\cdot) = \mu_2(\cdot) < \infty$, then $\mu_1 = \mu_2$. Hence, my idea is that it breaks down when we don't enforce finiteness. However, I'm struggling coming up with a counterexample of this specifically. Any ideas?
Let $\#$ be the counting measure on the space $(\mathbb{R}^n, \mathcal{B}^n)$, where $\mathcal{B}^n$ is the Borel $\sigma$-algebra. Just take $\mu_1= \#$ and $\mu_2= 2 \cdot \#$.
It is clear that $\mu_1$, $\mu_2$ are measures on the space $(\mathbb{R}^n, \mathcal{B}^n)$, $$\mu_1\left(\prod_{i=1}^n (x_i, y_i]\right) = \mu_2\left(\prod_{i=1}^n (x_i, y_i]\right) = +\infty$$ if, for all $i\in\{1, \cdots n\}$, $x_i < y_i \in \mathbb{R}$, and $$\mu_1\left(\prod_{i=1}^n (x_i, y_i]\right) = \mu_2\left(\prod_{i=1}^n (x_i, y_i]\right) = 0$$ if, there is an $i\in\{1, \cdots n\}$ such that $x_i = y_i \in \mathbb{R}$. But $\mu_1 \neq \mu_2$.