Let $(X,\Sigma,\mu)$ be a finite measure space with group structure $\cdot$ such that for each $x \in X$ and $S \in \Sigma$, we have $x \cdot S \in \Sigma$ and $\mu(x \cdot S)=\mu(S)$. Suppose $f$ is a function $X \to \mathbb{R}$, both bounded and $\Sigma$-measurable. Then for each $x \in X$, $$\int f(x \cdot -) \text{d}\mu=\int f \text{d}\mu.$$
I realize that by $f$'s properties and the Dominated Convergence Theorem, we may reduce the question to simple functions (right?). So if $f$ is a simple function, $f=\sum_i c_i \chi_{A_i}$, and $x \in X$, we have $$\int f \text{d}\mu = \sum_i c_i \mu(A_i) = \sum_i c_i \mu(x \cdot A_i) = \int f(x \cdot -) \text{d}\mu.$$
The problem is, how exactly do I justify the third equality above (that is, we are integrating a composition of functions on the RHS; how exactly do I turn it into a straight simple function in order to apply definitions, or something along those lines?)?
I think I now realize what I actually needed to do: $$\int f \text{d}\mu = \sum_i c_i \mu(A_i) = \sum_i c_i \mu(x^{-1} \cdot A_i) = \int \sum_i c_i\chi_{x^{-1} \cdot A_i} \text{d}\mu = \int f(x \cdot -) \text{d}\mu,$$ since for each $x'\in X$, $$f(x \cdot x') = \sum_i c_i \chi_{A_i}(x \cdot x') = \sum_i c_i \chi_{x^{-1} \cdot A_i}(x').$$