Compact Haar measure is bi-invariant

For fixed $h\in G$ define ${\mu }_h(E):=\mu (Eh)$; since right translation is a homeomorphism and $\mu$ is a regular Borel probability, ${\mu }_h$ is again a regular Borel probability, and it is left invariant: ${\mu }_h(gE)=\mu (gEh)=\mu (Eh)={\mu }_h(E)$ for all $g$, because $\mu$ is left invariant.

The measure ${\mu }_h$ is a regular left-invariant Borel probability by step 1.1, so the uniqueness among left-invariant regular Borel probabilities in [L1] gives ${\mu }_h=\mu$ for every $h$. Thus $\mu (Eh)=\mu (E)$ for all Borel $E$ and $h$: normalized Haar measure on a compact Lie group is right invariant as well as left invariant.

Hence the statement of this item is false: the correct conclusion is bi-invariance, and the uniqueness argument above shows that no example of a compact Lie group with a merely left-invariant normalized Haar measure exists. The trivial group is included: its sole probability measure is the point mass, invariant on both sides.