Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product.
Let $\mathsf{Vol}$ be the D1738: Euclidean real volume function on $\mathbb{R}^N$.

Let $x \in \mathbb{R}^N$. If $\varepsilon > 0$, then the open Euclidean real interval $\prod_{n = 1}^N (x_n - \varepsilon^{1/N} / 2, x_n + \varepsilon^{1/N} / 2)$ contains the singleton set $\{ x \}$. Thus, R1171: Isotonicity of Euclidean volume function implies \begin{equation} \begin{split} \mathsf{Vol} (\{ x \}) & \leq \mathsf{Vol} \Big( \prod_{n = 1}^N (x_n - \varepsilon^{1/N} / 2, x_n + \varepsilon^{1/N} / 2) \Big) \\ & = \prod_{n = 1}^N |x_n + \varepsilon^{1/N} / 2 - (x_n - \varepsilon^{1/N} / 2)| \\ & = \prod_{n = 1}^N |\varepsilon^{1/N}| \\ & = \prod_{n = 1}^N \varepsilon^{1/N} \\ & = \varepsilon \end{split} \end{equation} Since $\varepsilon > 0$ was arbitrary, we may extend this inequality to the limit on the right hand side to deduce $\mathsf{Vol}(\{ x \}) \leq 0$. Since $\mathsf{Vol}(\{ x \})$ is a nonnegative quantity, result R1043: Equality from two inequalities for real numbers states that $\mathsf{Vol}(\{ x \})$ equals $0$. $\square$