Lebesgue Measure is Diffuse

Theorem

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

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

Proof

A singleton $\set {\mathbf x} \subseteq \R^n$ is seen to be closed by combining:

Euclidean Space is Complete Metric Space

Metric Space is Hausdorff

Corollary to Compact Subspace of Hausdorff Space is Closed

Hence by Closed Set Measurable in Borel Sigma-Algebra:

$\set {\mathbf x} \in \map \BB {\R^n}$

where $\map \BB {\R^n}$ is the Borel $\sigma$-algebra on $\R^n$.

Write $\mathbf x + \epsilon = \tuple {x_1 + \epsilon, \ldots, x_n + \epsilon}$ for $\epsilon > 0$.

Then:

$\ds \set {\mathbf x} = \bigcap_{m \mathop \in \N} \horectr {\mathbf x} {\mathbf x + \frac 1 m}$

where $\horectr {\mathbf x} {\mathbf x + \dfrac 1 m}$ is a half-open $n$-rectangle.

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

$\ds \map {\lambda^n} {\horectr {\mathbf x} {\mathbf x + \frac 1 m} } = \prod_{i \mathop = 1}^n \frac 1 m = m^{-n}$

From Characterization of Measures, it follows that:

$\ds \map {\lambda^n} {\set {\mathbf x} } = \lim_{m \mathop \to \infty} m^{-n}$

which equals $0$ from Sequence of Powers of Reciprocals is Null Sequence.

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

$\map {\lambda^n} {\set {\mathbf x} } = 0$

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

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $11 \ \text{(i)}$
• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 6$: Problem $5 \ \text{(i)}$