is the limit of a specific simple function integrable?

Question

Let $f$ be a non-negative integrable function from some measure space $(X,\mathcal{A},\mu)$ to $\mathbb{R}$, equiped with the Borel-$\sigma$-algebra and the Lebesgue measure $\lambda$. Define for $r > 1$: $$f_r := \sum_{n=-\infty}^\infty r^n _{A_n},$$ where $A_n = \{x \in \mathbb{R}; r^n \leq f(x) < r^{n+1}\}.$

Show $ f_r ≤ f$, for all $r > 1$ and for $r ↓ 1$, $f_r \to f$ almost everywhere.

$ f_r ≤ f$ for all $r > 1$ follows directly from the definition of $A_n$.

My guess for the second question is that it has everything to do with the $A_n$. The set of all $A_n$ form an partition of $\mathbb{R}^{+}$, and as $r ↓ 1$, gets finer and finer. Is my reasoning correct and how do I form this argument mathematically?

Answer 1

First of all, note that $$\tag{1}\lim_{r\to1}(r^{n+1}-r^n)=0.$$ Let $x\in\mathbb{R} = \bigcup_{n=-\infty}^\infty A_n$. Then there is a $n\in\mathbb{Z}$ such that $x\in A_n$ and $x\notin A_m$ for $m\neq n$. This way $f_{n}(x)=r^n.$ We want to prove that $$\tag{2}\lim_{r\to1} f_n(x)=f(x).$$ Let $\epsilon >0$ arbitrary. It follows by (1) that there exists $\delta>0$ such that $$\tag{3}0<|r-1|<\delta \Rightarrow r^{n+1}-r^n=|r^{n+1}-r^n|<\epsilon.$$ On the other hand, since $x\in A_n$, $f_n(x)=r^n\leq f(x)

$$\tag{4}0<|r-1|<\delta \Rightarrow |f(x)-f_n(x)| = f(x)-r^n \leq r^{n+1}-r^n<\epsilon.$$

Now (2) follows from (4).