Let $\mu$ be a Borel measure on a topological space $X$, and let $E \subseteq X$ be Borel. Let $\phi$ be a homeomorphism of $X$, and let $\lambda$ be the measure given by $\lambda(A) = \mu(\phi(A))$. If $f: E \rightarrow \mathbb{C}$ is measurable and integrable, then so is $f \circ \phi: \phi^{-1}E \rightarrow \mathbb{C}$, and
$$\int\limits_E f \space d \mu= \int\limits_{\phi^{-1}E} f \circ \phi \space d\lambda$$
For example, if $f$ is a simple function, say $f = \sum\limits_{i=1}^m c_i \chi_{E_i}$ for $E_i \subseteq E$ disjoint and Borel, then $f \circ \phi: \phi^{-1}E \rightarrow \mathbb{C}$ is the function $\sum\limits_{i=1}^m c_i \chi_{\phi^{-1}E_i}$, and
$$\int\limits_{\phi^{-1}E} f \circ \phi \space d \lambda = \sum\limits_{i=1}^m c_i \lambda(\phi^{-1}E_i) = \sum\limits_{i=1}^m c_i\mu(E_i) = \int\limits_E f \space d \mu$$
Now let $G$ be a locally compact topological group, $\mu$ a left Haar measure on $G$, and $\phi$ the topological group isomorphism $x \mapsto x_0xx_0^{-1}$ for a fixed $x_0 \in G$. Define $\lambda = \mu \circ \phi$. It is easy to see that $\lambda(E) = \mu(x_0Ex_0^{-1}) = \mu(Ex_0^{-1})$ for all Borel sets $E$, and that $\lambda$ is also a left Haar measure. Therefore, there exists a constant $\Delta(x_0) = \Delta > 0$ such that $\lambda = \Delta \mu$, or in other words
$$\mu(Ex_0^{-1}) = \Delta \mu(E)$$
for all $E$ Borel. Now, for $f: G \rightarrow \mathbb{C}$ integrable, we should have
$$\int\limits_G f d \mu = \int\limits_{\phi^{-1}G} f \circ \phi \space d \lambda = \int\limits_G f \circ \phi \space d\lambda = \Delta \int\limits_G f \circ \phi \space d\mu$$
Or in other words,
$$\int\limits_G f(x)dx = \Delta(x_0) \int\limits_G f(x_0xx_0^{-1}) dx$$
However, the notes I'm reading here (http://www.math.toronto.edu/murnaghan/courses/mat1196/rnotes.pdf) page 19, have $\Delta(x_0)$ on the other side of the equation. Am I making a mistake somewhere?
I did not deeply delve in the topic, but, as far as I understood, the error is in the notes. For instance, if $f=\chi_S$ then
$$\int_G f(gg_0^{-1})d_\ell g = \int_G \chi_S(gg_0^{-1})d_\ell g =\int_G \chi_{Sg_0}(g)d_\ell g=\mu_\ell(Sg_0)=\mu_\ell(S)/\Delta(g_0).$$