Proof of Countable Sets are Null for an Atomless Measure, One-Point Sets are Lebesgue Null, and an Absolutely Continuous Measure is Atomless

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Let $\iota$ be the canonical map from $\mathbb{N}$ to $\mathbb{R}$, positive with positive inverse by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; this is the only use of that symbol here. For $x\in\mathbb{R}^{q}$ the one-point set $\{x\}$ belongs to $\mathcal{B}(\mathbb{R}^{q})$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (v).

Step 1 (A countable set is Borel and its measure is bounded by the sum of the point masses). Let $\alpha$ be a measure on $(\mathbb{R}^{q},\mathcal{B}(\mathbb{R}^{q}))$ with $\alpha(\{x\})=0$ for every $x\in\mathbb{R}^{q}$, and let $A\subseteq\mathbb{R}^{q}$ be countable. If $A=\varnothing$, then $A\in\mathcal{B}(\mathbb{R}^{q})$ and $\alpha(A)=0$ by Measure, Measure Space, and Probability Measure. Otherwise, by Countable Set there is a sequence $(a_{m})_{m\in\mathbb{N}}$ whose set of terms is $A$, so that

$$A=\bigcup_{m\in\mathbb{N}}\{a_{m}\},$$

which belongs to $\mathcal{B}(\mathbb{R}^{q})$, a $\sigma$-algebra being closed under countable unions. By claim 4 of Basic Properties of a Measure,

$$\alpha(A)\le\sum_{m\in\mathbb{N}}\alpha(\{a_{m}\})=0,$$

the series having every term $0$; and $0\le\alpha(A)$ by Measure, Measure Space, and Probability Measure, so $\alpha(A)=0$.

Step 2 (Claim 1). If $\mu$ is atomless then $\mu(\{x\})=0$ for every $x$, so Step 1, applied with $\alpha=\mu$, gives $A\in\mathcal{B}(\mathbb{R}^{q})$ and $\mu(A)=0$.

Step 3 (Claim 2). Let $x\in\mathbb{R}^{q}$ and $j\in\mathbb{N}$. By claims 2 and 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square one has $d_{E}(x,x)=\lVert x-x\rVert=0$, so $x\in\bar{B}(x,\iota(j)^{-1})$ and $\{x\}\subseteq\bar{B}(x,\iota(j)^{-1})$. By claim 3 of The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ and claim 2 of Basic Properties of a Measure,

$$0\le\lambda_{q}(\{x\})\le\kappa_{q}\bigl(\iota(j)^{-1}\bigr)^{q},$$

where $\kappa_{q}$ is the real number of claim 2 of that lemma; in particular $\lambda_{q}(\{x\})$ is a real number. The sequence $(\iota(j)^{-1})_{j\in\mathbb{N}}$ converges to $0$: given a positive real $\varepsilon$, claim 3 of The Archimedean Property of the Real Numbers provides $j_{0}\in\mathbb{N}$ with $\iota(j_{0})^{-1}<\varepsilon$, and every $j\ge j_{0}$ satisfies $\iota(j_{0})\le\iota(j)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, hence $\iota(j)^{-1}\le\iota(j_{0})^{-1}<\varepsilon$ on multiplying by $\iota(j_{0})^{-1}\iota(j)^{-1}$, which is nonnegative by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field, using claim 5 of Elementary Arithmetic in an Ordered Field and claim 2 of Elementary Order Arithmetic in an Ordered Field. Consequently $\bigl(\kappa_{q}(\iota(j)^{-1})^{q}\bigr)_{j}$ converges to $0$ by claims 2 and 3 of Arithmetic of Limits of Real Sequences, the power being a finite product. Hence $\lambda_{q}(\{x\})\le\eta$ for every positive real $\eta$, and Comparison of Real Numbers with Arbitrary Positive Slack (iii) gives $\lambda_{q}(\{x\})=0$.

The second assertion of claim 2 now follows from Step 1, applied with $\alpha=\lambda_{q}$, which satisfies $\lambda_{q}(\{x\})=0$ for every $x$ by what has just been proved.

Step 4 (Claim 3). Suppose $\mu$ is absolutely continuous and let $x\in\mathbb{R}^{q}$. The set $\{x\}$ belongs to $\mathcal{B}(\mathbb{R}^{q})$ and satisfies $\lambda_{q}(\{x\})=0$ by claim 2, so Absolutely Continuous Probability Measure on Euclidean Space (i) gives $\mu(\{x\})=0$. As $x$ was arbitrary, $\mu$ is atomless.