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Do linear and group invariant functions allowed to go inside(?) integral operators?

I am working through the book *Geometric Deep Learning* (https://arxiv.org/abs/2104.13478) and have hit the following formula (Chapter 3.5, page 27). 

> If $f$ is linear and $\mathfrak G$-invariant and $\mu(\mathfrak g)$ is Haar measure of the group $\mathfrak G$, then for all $x\in \mathcal{X}(\Omega)$,
> $$f(x) = \frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G} f(\mathfrak g.x)d\mu(\mathfrak g) = f \left(\frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G}(\mathfrak g.x)d\mu(\mathfrak g)\right)$$
> which indicates that $F$ only depends on $x$ through the $\mathfrak G$-average $Ax= \frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G} f(\mathfrak g.x)d\mu(\mathfrak g)$.

Two questions.

1. It doesn't seem obvious to me that the function $f$ being linear and $\mathfrak G$-invariant allows the integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ to just go inside(?) the function $f$. The authors gave no explation as to how it is done. How is the formula derived? 

2. I don't know how this operation of integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ 'going inside' the function $f$ is called. Would 'commute' be appropriate term? (For example, the function $f$ and integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ commute with each other.)