Are every measure on $\mathbb{R}^{n}$ borel and/or regular? I saw the above question in an exam paper and I am not sure how to even start. The question is true for the case of Lebesgue measure but I am not sure for arbitrary measures. I have tried looking at non-measurable sets but I can't seem to find the solution.
I have not been able to find any proof or counter examples anywhere.
 A: Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^n$, and $\mathcal{L}$ the Lebesgue $\sigma$-algebra.
Take a Vitali set $V\subseteq[0,1]^n$ (i.e., a set of representatives of $\mathbb{R}^n/\mathbb{Q}^n$), which is non-Lebesgue-measurable for $\lambda$ and has cardinality $|V|=|\mathbb{R}^n|=\mathfrak{c}$. Take a nonempty open set $B\subseteq(2,3)^n$, which thus has same cardinality as $V$, and take a bijection $f:V\to B$.
Define a bijection $T:\mathbb{R}^n\to\mathbb{R}^n$ by setting $T(x)=f(x)$ if $x\in $, $T(x)=f^{-1}(x)$ is $x\in B$, and $T(x)=x$ for $x\in\mathbb{R}^n\setminus(V\cup B)$.
Let $(\mathcal{B},\mu)=(T(\mathcal{L}),T_*\lambda)$, i.e., define a $\sigma$-algebra $\mathcal{B}$ such that $A\in\mathcal{B}$ if and only if $T^{-1}(A)\in\mathcal{L}$, and $\mu(A)=\lambda(T^{-1}(A))$ in this case. Let's show that $B$ is not measurable for $(\mathcal{B},\mu)$.
A set $A\subseteq\mathbb{R}^n$ is (Caratheodory-)measurable for $(\mathcal{B},\mu)$ if and only if for every $E\in\mathcal{B}$,
$$\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)$$
or equivalently, for every $F\in\mathcal{L}$
$$\lambda^*(T^{-1}(A))=\lambda^*(T^{-1}(A)\cap F)+\lambda^*(T^{-1}(A)\setminus F)$$
Since $T^{-1}(B)=V$ is not Lebesgue-measurable, and hence also not $\lambda$-measurable in the sense of Caratheodory (this follows from the fact that the Lebesgue $\sigma$-algebra is itself the one of Caratheodory-measurable sets for the Borel $\sigma$-algebra), then $B$ is not measurable for $\mu$.

As for the second part, the counting measure on $\mathcal{P}(\mathbb{R}^n)$ is not (outer) regular.
In fact, you can separate $\mathbb{R}^n$ in two parts, and apply each previous procedure to each part, and obtain a measure on $\mathbb{R}^n$ which is at the same time non-regular and for which not every Borel set is measurable.
