Is the Borel sigma algebra complete under some measure?

A measure space $$(X,F,\mu)$$ is complete if whenever $$\mu(E)=0$$, every subset of $$E$$ is an element of $$F$$. Every incomplete measure space has a completion, which extends the sigma algebra to more sets and extends the measure to those sets. Most famously, the Borel algebra on $$\mathbb{R}$$ is incomplete with respect to Lebesgue measure, but it can be extended to a bigger sigma algebra, the Lebesgue signma algebra, which is complete with respect to Lebesgue measure.

But question is, does there exist a measure under which the Borel sigma algebra on $$\mathbb{R}$$ is complete? Or is it incomplete under every measure?

The Borel $$\sigma$$-algebra on any space is complete with respect to counting measure, since the only null set for counting measure is the empty set.
However, the Borel $$\sigma$$-algebra on $$\mathbb{R}$$ is not complete with respect to any $$\sigma$$-finite measure. Note that $$\mathbb{R}$$ is Borel-isomorphic to $$\mathbb{R}^2$$ (see Are the measurable spaces $$(\mathbb{R}^n, Bor(\mathbb{R}^n))$$ and $$(\mathbb{R}^m, Bor(\mathbb{R}^m))$$ isomorphic for $$n\neq m$$, for instance), so let us work with $$\mathbb{R}^2$$ instead of $$\mathbb{R}$$. So suppose $$\mu$$ is a $$\sigma$$-finite Borel measure on $$\mathbb{R}^2$$. Note then the sets $$\mathbb{R}\times\{t\}$$ as $$t$$ ranges over $$\mathbb{R}$$ are disjoint, and so by $$\sigma$$-finiteness of $$\mu$$ we must have $$\mu(\mathbb{R}\times\{t\})=0$$ for all but countably many $$t\in\mathbb{R}$$. Since not every subset of $$\mathbb{R}\times\{t\}$$ is Borel, this measure space is not complete.