Lebesgue Measure is Diffuse

Theorem
Let $\lambda^n$ be Lebesgue measure on $\R^n$.

Then $\lambda^n$ is a diffuse measure.

Proof
Any singleton $\left\{{\mathbf x}\right\} \subseteq \R^n$ is closed by combining:


 * Euclidean Space is Complete Metric Space
 * Metric Space is Hausdorff
 * Corollary to Compact Subspace of Hausdorff Space is Closed

Whence by Closed Set Measurable in Borel Sigma-Algebra, $\left\{{\mathbf x}\right\} \in \mathcal B \left({\R^n}\right)$.

Here, $\mathcal B \left({\R^n}\right)$ is the Borel $\sigma$-algebra on $\R^n$.

Write $\mathbf x + \epsilon = \left({x_1 + \epsilon, \ldots, x_n + \epsilon}\right)$ for $\epsilon > 0$.

Then:


 * $\displaystyle \left\{{\mathbf x}\right\} = \bigcap_{m \mathop \in \N} \left[\left[{\mathbf x \,.\,.\, \mathbf x + \frac 1 m}\right)\right)$

where $\left[\left[{\mathbf x \,.\,.\, \mathbf x + \dfrac 1 m}\right)\right)$ is a half-open $n$-rectangle.

By definition of Lebesgue measure, we have (for all $m \in \N$):


 * $\displaystyle \lambda^n \left({\left[\left[{\mathbf x \,.\,.\, \mathbf x + \frac 1 m}\right)\right)}\right) = \prod_{i \mathop = 1}^n \frac 1 m = m^{-n}$

From Characterization of Measures, it follows that:


 * $\displaystyle \lambda^n \left({\left\{{\mathbf x}\right\}}\right) = \lim_{m \to \infty} m^{-n}$

which equals $0$ from Power of Reciprocal.

Therefore, for each $\mathbf x \in \R^n$:


 * $\lambda^n \left({\left\{{\mathbf x}\right\}}\right) = 0$

that is, $\lambda^n$ is a diffuse measure.