Here, $\mathcal{C}(G, \mathbb{R})$ denotes the continuous functions from a topological group $G$ to $\mathbb{R}$.
I'm taking a course on the representation theory of groups and have moved onto a section about topological groups, and we have defined the Haar integral as a linear map
$$\smallint_G: \mathcal{C}(G, \mathbb{R}) \to \mathbb{R}$$
with the properties
- $\int_G \mathbf{1}_G = 1$ (so $\int_G$ is normalised so that the total value is $1$)
- $\int_G f(xg) dg = \int_G f(g)dg = \int_G f(gx) dg$ for all $x \in G$ (so $\int_G$ is translation-invariant)
- We write $\int_G f(g)dg = \int_G f$ and $\int_G f(xg) dg$ means apply $\int_G$ to $g \mapsto f(xg) \in \mathcal{C}(G, \mathbb{R})$.
- $\int_G f \geq 0$ if $f(g) \geq 0$ for all $g \in G$ (positivity)
I've only self-studied measure theory, and am wondering how these properties of the Haar integral relate to the Haar measure. For example, I know the Haar measure is translation invariant, like the Haar integral, but (how do) the other properties of the Haar integral correspond with other properties of the Haar measure?
Best Answer
Integral and measure are basically the same thing, in my opinion.
Imagine $\mathbb{R}^2$... then you say:
Those "many sets" are the measurable sets... the $\sigma$-algebra. The rectangles are what is called a semi-ring. So, in order to have the measure, all you need is to know the size of the rectangles. This is Carathéodory's extension theorem. For some reason, people get satisfied when they can measure the sets in some $\sigma$-algebra.
However, one might argue that what you really want is to integrate things. Then, you realize that if you know the measure, you can integrate "positive measurable functions" ($f \geq 0$). In a sense, this is an extension theorem! Basically you started with the integral of some basic functions... the indicator functions: \begin{align*} I_B: \Omega &\rightarrow \mathbb{R} \\ w &\mapsto \begin{cases} 1,& w \in B \\ 0, &\text{otherwise} \end{cases} \end{align*} The integral of such functions is just the measure: \begin{equation*} \int I_B\,\mathrm{d}\mu = \mu(B). \end{equation*} So, in this sense, the integral is the extension of the measure from indicator to step functions, to positive measurable functions, to integrable functions.
But you do not need to start with functions like $I_B$. People might want to start with continuous functions. They define the integral (a continuous linear functional) and they extend it to include other functions. If you can then, integrate the indicator functions ($I_B$), you have recovered the measure. (from this same measure, can you recover the integral you started with??? what does this integral have to satisfy so that both notions coincide???...)
So, integral and measure are basically the same thing viewed in two different ways.