I've looking at the following proof of the uniqueness of Haar measure on compact groups. I think I understand every step in the proof but I don't see where regularity is used. I see that the both fact that the measures are translation invariant and that $H(G)=1=H'(G)$ (in particular to guarantee that Fubini's theorem holds) are used extensively, but regularity doesn't seem to be necessary for uniqueness. Is this the case or am I missing something?
Let $H'$ be a measure with the same properties as $H$ and let $f\in C(G)$. Then \begin{equation*} \begin{split} \int f(y)dH'(y)&=\int H(G)f(y)dH'(y)=\int(\int f(y)dH'(y))dH(y)\\ &=\int(\int f(xy)dH'(y))dH(x)=\int(\int f(xy)dH(x))dH'(y)\\ &=\int(\int f(x)dH(x))dH'(y)=\int H(G)f(x)dH(x)=\int fdH(x) \end{split} \end{equation*}
Therefore, $H'=H$.
By the way, to show uniqueness shouldn't it be enough to show that the above equalities hold for indicator functions? From a set theoretic perspective it should be enough since two functions are defined to be equal if they are equal pointwise.
PS: I'm sorry about the notation I used. I realize it's not the best but it's the one I've been using.