Calculus notation in Haar measure. What is really going on? Let $G$ be a locally compact group with left Haar measure $\mu$.  If $f: G \rightarrow \mathbb{C}$ is an integrable function, and $E \subseteq G$, the notation 
$$\int_E f(g)dg$$
is commonly used instead of $$\int\limits_E f d\mu$$
It is also common to write things like $\int\limits_E f(g)d(gh^{-1})$ for a fixed $h \in H$.  This seems to be somewhat like u-substitution, although there should be some formal measure theory behind it. Would anyone be willing to explain or give a reference which explains the measure-theoretic principles behind the use of such notation?
 A: Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.


*

*On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.

*For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=\lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $\lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$. 
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $\cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance. 
3c. The canonical example (from wikipedia) is the topological group $(\mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.


*Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $\int_G f dm = \int_G f(x) dm(x)$. Having determined a unique measure, we then write $\int_G f(x) dx$ to mean $\int_G f(x) dm(x)$.


The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
