Proof of The Support of a Borel Measure is Closed, and Carries Full Measure on a Separable Space

Each result cited below is universally quantified over the data in its own statement.

Claim 1. Let $x\in X\setminus\operatorname{supp}\mu$. By Support of a Borel Measure on a Metric Space (i) there is $r\in\mathbb{R}$ with $0<r$ for which $0<\mu(B_{d}(x,r))$ fails; as $0$ is the least element of $[0,\infty]$ in the order fixed in Measure, Measure Space, and Probability Measure, this means $\mu(B_{d}(x,r))=0$.

Let $y\in B_{d}(x,r)$, so that $d(x,y)<r$ by Open Ball in a Metric Space, and put $s=r-d(x,y)$, a positive real number. If $w\in B_{d}(y,s)$ then the triangle inequality of Metric Space gives

$$d(x,w)\le d(x,y)+d(y,w)<d(x,y)+s=r ,$$

the middle step by strict compatibility of the order with addition and mixed transitivity (claims 1 and 2 of Elementary Order Arithmetic in an Ordered Field); hence $w\in B_{d}(x,r)$. Thus $B_{d}(y,s)\subseteq B_{d}(x,r)$, and claim 2 of Basic Properties of a Measure gives $\mu(B_{d}(y,s))\le\mu(B_{d}(x,r))=0$, so $\mu(B_{d}(y,s))=0$ and $y\notin\operatorname{supp}\mu$.

Therefore $B_{d}(x,r)\subseteq X\setminus\operatorname{supp}\mu$. Since $x\in X\setminus\operatorname{supp}\mu$ was arbitrary, Open Subset of a Metric Space gives $X\setminus\operatorname{supp}\mu\in\mathcal{T}_{d}$. Consequently $\operatorname{supp}\mu$, the complement of a member of $\mathcal{T}_{d}$, is closed in $(X,\mathcal{T}_{d})$; and $X\setminus\operatorname{supp}\mu$ belongs to $\mathcal{B}(X,d)$ by Borel Sigma-Algebra of a Metric Space, hence so does its complement $\operatorname{supp}\mu$ by σ-Algebra and Measurable Space.

Claim 2. Assume $(X,d)$ separable. If $X=\emptyset$ then $X\setminus\operatorname{supp}\mu=\emptyset$ and $\mu(\emptyset)=0$ by Measure, Measure Space, and Probability Measure, so assume $X\neq\emptyset$ and fix $x_{0}\in X$.

By Separable Metric Space there is a countable $D\subseteq X$ that is dense in $X$, that is $\operatorname{cl}_{X}(D)=X$. Applying claim 3 of Characterization of the Closure in a Metric Space by Open Balls to $x_{0}\in\operatorname{cl}_{X}(D)$ produces a point of $D$, so $D\ne\emptyset$ and, by Countable Set, $D$ is the set of terms of a sequence $(q_{j})_{j\in\mathbb{N}}$. The set $\mathbb{Q}$ of rational numbers is countable by The Integers and the Rational Numbers are Countable and nonempty, hence is the set of terms of a sequence $(t_{l})_{l\in\mathbb{N}}$.

For $j,l\in\mathbb{N}$ put $A_{j,l}=B_{d}(q_{j},t_{l})$ if $0<t_{l}$ and $\mu(B_{d}(q_{j},t_{l}))=0$, and $A_{j,l}=\emptyset$ otherwise; in either case $A_{j,l}\in\mathcal{B}(X,d)$ (an open ball is open by Open Ball in a Metric Space is Open, hence Borel by Borel Sigma-Algebra of a Metric Space) and $\mu(A_{j,l})=0$. Put $V_{j}=\bigcup_{l\in\mathbb{N}}A_{j,l}$ and $U=\bigcup_{j\in\mathbb{N}}V_{j}$. By countable subadditivity (claim 4 of Basic Properties of a Measure) $\mu(V_{j})\le\sum_{l\in\mathbb{N}}\mu(A_{j,l})=0$, the sum of a sequence with every term $0$ being $0$ by the conventions of Measure, Measure Space, and Probability Measure; the same lemma applied to $(V_{j})_{j\in\mathbb{N}}$ gives $\mu(U)=0$.

It remains to show $X\setminus\operatorname{supp}\mu\subseteq U$. Let $x\notin\operatorname{supp}\mu$ and, as in claim 1, take $r\in\mathbb{R}$ with $0<r$ and $\mu(B_{d}(x,r))=0$. By claim 8 of Elementary Order Arithmetic in an Ordered Field the number $r/2$, the product of $r$ with the inverse of $2=1+1$, is positive, so by The Rational Numbers are Dense in the Real Numbers there is $t\in\mathbb{Q}$ with $0<t<r/2$; choose $l\in\mathbb{N}$ with $t_{l}=t$. Since $\operatorname{cl}_{X}(D)=X$, claim 3 of Characterization of the Closure in a Metric Space by Open Balls gives $j\in\mathbb{N}$ with $d(x,q_{j})<t$, so $x\in B_{d}(q_{j},t)$ by the symmetry of $d$ (Metric Space). For $w\in B_{d}(q_{j},t)$ the triangle inequality gives

$$d(x,w)\le d(x,q_{j})+d(q_{j},w)<t+t=2t<r ,$$

using claims 1, 2 and 10 of Elementary Order Arithmetic in an Ordered Field for the two strict inequalities. Hence $B_{d}(q_{j},t)\subseteq B_{d}(x,r)$ and $\mu(B_{d}(q_{j},t))\le\mu(B_{d}(x,r))=0$ by claim 2 of Basic Properties of a Measure. Therefore $A_{j,l}=B_{d}(q_{j},t)$, and $x\in A_{j,l}\subseteq U$.

Both $X\setminus\operatorname{supp}\mu$ (by claim 1) and $U$ belong to $\mathcal{B}(X,d)$, so claim 2 of Basic Properties of a Measure gives $\mu(X\setminus\operatorname{supp}\mu)\le\mu(U)=0$, that is $\mu(X\setminus\operatorname{supp}\mu)=0$.