Limit of the integral of a measurable function

I'm working on the following problem.

Let $f \geq 0$ a measurable and integrable function on $(X,\mathcal{A},\mu)$ a measurable space, and for $r>1$ we define

$$f_r(x) = \sum_{n=-\infty}^{\infty} r^n \mathbf{1}_{A_{n,r}}(x)$$

where $A_{n,r} = \{x \in X: r^n \leq f(x) < r^{n+1}\}$

I want to show that $\int_X f_r \text{ } d\mu \rightarrow \int_X f \text{ } d\mu$ when $r \rightarrow 1$

So far I have managed to prove that

• $f_r$ is measurable $\forall r>1$
• $\int_X f_r \text{ } d\mu = \sum_{n=-\infty}^{\infty} r^n \mu({A_{n,r}})$

And in the way I have rewriten

$$f_r(x) = r^0A_{0,r} + \sum_{n=1}^{\infty}u_n + \sum_{n=1}^{\infty}v_n $$

where $(u_n)_{n \in \mathbb{N}} = r^n \mathbf{1}_{A_{n,r}}(x)$, $(v_n)_{n \in \mathbb{N}} = r^{-n} \mathbf{1}_{A_{-n,r}}(x)$

Now I would like to use a convergence theorem (I was thinking about using the dominated convergence theorem, the monotone convergence theorem or a mix or both) to get to the final result, but this is where I've gotten stuck.

It also confuses me that the limit I'm trying to calculate takes $r \rightarrow 1$

Thanks in advance for any help!

Lets first show that $s(r):=\sum_{n=0}^{\infty} r^n \mathbf{1}_{A_{n,r}}$ satisfies $s(r)\to f_1:=f1_{\{f\geq 1\}}$ pointwise as $r\downarrow 1$. Define the sequence of simple functions $s_k:=\sum_{n=0}^{k} r^n \mathbf{1}_{A_{n,r}} $ so that $s_k\uparrow s$ pointwise.

Note that $s_k(r)\leq f_1$ for each $k$ so that $s(r)\leq f_1 $. Moreover we have $$|f_1-s_k(r)|\leq (r-1) \sum_{n=0}^{k} r^n \mathbf{1}_{A_{n,r}} + f1_{f\geq r^{k+1}} \to (r-1) \sum_{n=0}^{\infty} r^n \mathbf{1}_{A_{n,r}} $$

pointwise as $k\to\infty$. This means that $|f_1-s(r)|\leq (r-1) \sum_{n=0}^{\infty} r^n \mathbf{1}_{A_{n,r}}$ pointwise. Now let $r\downarrow 1$ to get $s(r)\to f_1$ pointwise as $r\downarrow 1$. Since $f$ is assumed integrable we can use the DCT on an arbitrary sequence $r_n\downarrow 1$ to obtain $\int s(r_n) \to \int f_1$ as $n\to \infty$. As $(r_n)$ was arbitrary this implies $\int s(r) \to \int f_1$ as $r\downarrow 1$.

Can you do something similar for $f1_{\{f< 1\}}$?