I am reading about ergodic theory and have a question. Let $f : M \rightarrow M$ be a measurable mapping (the space where the mapping happens is not important right now). A measure $\mu$ is invariant under $f$ if $$\mu (E) = \mu (f^{-1} (E))$$ for all measurable subsets $E \subset M.$ The author says that you can think of this condition as "the probability that a randomly chosen point is in $E$ is the same as the probability that its image is in $E$." However, why wouldn't that mean that $$\mu(E) = \mu (f(E))?$$
What is so special about $\mu (E) = \mu (f^{-1} (E))$? What does this mean intuitvely?
Let's take a closer look at this. By definition $$ x\in f^{-1}(E) \iff f(x) \in E.$$ If the mass $\mu(E)$ is the same as $\mu(f^{-1}(E))$, then "probability that randomly chosen point is in $E$ equals probability that this same points image is in $E$". This is a bit informal. Instead of looking at some hard to define random points, let's see more verbose notation of this situation:
\begin{aligned} \mu(E) &=\mu(f^{-1}(E)) \iff\\ \mu(\{x\in X: x\in E\})& =\mu(\{x\in X: f(x)\in E\}). \end{aligned}
It may be unnatural at first, but as you can see intuition given by this author checks out.