Non-measurable sets on groups from translation invariance

The proof for the reals can be generalized to any non-discrete locally compact group $G$. We let $K \subset G$ be any compact set with positive Haar measure $\lambda(K) > 0$ (e.g., $K = [0, 1]$ when $G = \mathbb R$), and we let $\Lambda < G$ be any subgroup such that $\Lambda \cap KK^{-1}$ is countably infinite (e.g., $\Lambda = \mathbb Q$ when $G = \mathbb R$). We define an equivalence relation on $G$ by the $\Lambda$-orbits coming from left multiplication and we let $V \subset K$ be a set containing exactly one representative of each equivalence class that intersects non-trivially with $\Lambda K$.

We then have $K \subset (\Lambda \cap K K^{-1}) V \subset K K^{-1} K$. The first inclusion here follows from the fact that if $k \in K$, then we may find $t \in \Lambda$ such that $tk \in V \subset K$. It then follows that $t = (tk) k^{-1} \in K K^{-1}$ and hence $k \in (\Lambda \cap K K^{-1})V$.

If we were able to extend the Haar measure to a countably additive left-invariant measure defined on all subsets of $G$, then we would have $\lambda(( \Lambda \cap K K^{-1} ) V) = \sum_{t \in \Lambda \cap K K^{-1}} \lambda(V) \in \{ 0, \infty \}$, but this would then contradict the inequalities $$ 0 < \lambda(K) \leq \lambda( ( \Lambda \cap K K^{-1} ) V ) \leq \lambda(K K^{-1} K) < \infty. $$

Not an answer (does not use translation invariance)

Another non-measurable set, which may generalize more easily, is the Bernstein set ... That is, a set $E$ such that for every uncountable closed set A, we have $A \cap E \ne \varnothing$ and $A \setminus E \ne \varnothing$ .

[With AC] we can prove that any uncountable Polish space admits a Bernstein set (indeed, $\mathfrak c$ disjoint Bernstein sets). If $\mu$ is any atomless Borel measure on an uncountable Polish space, then a Bernstein set is not $\mu$-measurable.