This is an exercise on page 158 of Conway's text A course in functional analysis:
Let $G$ be a locally compact group. If $m$ is a regular Borel measure on $G$, show that any two of the following properties imply the third:
- $m (\Delta g)= m(\Delta)$ for every Borel set $\Delta$ and every $g$ in $G$;
- $m(g\Delta) = m(\Delta)$ for every Borel set $\Delta$ and every $g$ in $G$;
- $m(\Delta) = m(\Delta^{-1})$ for every Borel set $\Delta$.
The proof for 1&3 $\Rightarrow$2 is quite simple $m(g\Delta)=m(\Delta^{-1}g^{-1})=m(\Delta^{-1})=m(\Delta)$. Similarly, we can prove 2&3 $\Rightarrow$ 1. But how to prove 1&2 $\Rightarrow$ 3? How to obtain $\Delta^{-1}$ from $\Delta$?
I assume that you know that a (left-invariant) Haar measure is unique up to constants.
If 1) and 2) hold, define $m'(B) := m(B^{-1})$. It is not hard to check that $m'$ is a regular Borel measure. Furthermore, $m'(gB) = m((gB)^{-1}) = m(B^{-1} g^{-1}) = m(B^{-1}) = m'(B)$, so that $m$ and $m'$ are both left-invariant regular Borel measures on $G$, and hence $m' = c \, m$ for some $c \in [0,\infty]$ (note that $m \equiv 0$ if and only if $m' \equiv 0$).
The claim is clear if $m = 0$. Otherwise, by regularity, there is a compact set $K \subset G$ of positive measure. Set $K' := K \cup K^{-1}$ and note $m(K') = m'(K')$, so that necessarily $c=1$; that is, $m = m'$.