How do I show that $P(X\in \text{Supp(X)})=1$ For a random vector X I want to show that  $P(X\in \text{Supp(X)})=1$ for a random vector X on $\mathbb{R}^n$ given the following definition of the support:
$$\text{Supp(X)}=\{x\in\mathbb{R^n} : P(X\in B_\epsilon(x))>0 \forall \epsilon>0\}$$
My initial thoughts were to show that $P(X\in\text{Supp(X)}^C)=0$ by firstly showing that the complement of the support is the union of all open null-sets and then using sub-additivity, but I'm unsure how exactly to go about this part (or if this is even the right approach).
 A: As in your OP, suppose $X:(\Omega,\mathscr{F},\mu)\rightarrow(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$ is a measurable function. Here $(\Omega,\mathscr{F},\mu)$ is a measure space (not necessarily a probability space). The support of $X$, or rather, the support if the  measure $\mu_X=\mu\circ X^{-1}$ on $\mathscr{B}(\mathbb{R}^n)$ induced by $X$ is defined as
$$\operatorname{supp}(\mu_X)=\{x\in \mathbb{R}^n: \mu(B(x;\delta))>0, \,\text{for all}\, \delta>0\}$$

Lemma: $\operatorname{supp}(\mu)$ is a closed set. Furthermore, there is a collection $\mathcal{U}$ of open balls $B\subset U:=\mathbb{R}^n\setminus\operatorname{supp}(\mu)$ with  $\mu(B)=0$ such that $U=\bigcup\{B:B\in\mathcal{U}\}$.

Proof: If $x\in U=\mathbb{R}^n\setminus\operatorname{supp}(\mu_X)$, then by definition of $\operatorname{\mu}$, there is $\delta_x>0$ such that $\mu(B(x;\delta_x))=0$.
Claim: $B(x;\delta_x)\subset U$: suppose $z\in B(x;\delta_x)$ and let $\delta'=\min(d(x,z),\delta_x-d(x,z))$. Then $B(z;\delta')\subset B(x;\delta_x)$ and so, $\mu(B(z;\delta')\leq \mu(B(x;\delta_x)=0$. Again, by definition of $\operatorname{\mu}$ we conclude that $z\in U$; consequently, $B(x;\delta_x)\subset U$.
It follows that the collection $\{B(x_x;\delta_x):x\in U\}$ is an open cover of $U$. $\Box$.
To conclude the problem notice that since the Euclidean space $\mathbb{R}^n$ is a separable metric space, $U$ admits a countable subcover $\mathcal{V}=\{B_n:n\in\mathbb{N}\}\subset\mathcal{U}$ ((Lindelöf's lemma). The conclusion follows immediately:
$$\mu(U)\leq\mu\big(\bigcup_n B_n\big)\leq\sum_n\mu\big(B_n\big)=0$$
