Show $\lim_{n\to \infty}\sum_{k=1}^{n} r_{k}\chi _{A_{k}} = f$

Question

My Attemp: Let $(X,\mathcal{M})$ a measurable space and $:X→[0,\infty]$ measurable and , $(r_{n})_{n \in {\mathbb{N}}}$ is a sequence $(0,\infty)$ such that $r_{n} \to 0$ and $\sum_{n=1}^{\infty} r_{n}= \infty$ , where \begin{equation*} \begin{split} A_1:=\{x\in X: r_1 \leq f(x) \} \hspace{3ex} & \text{and} \hspace{3ex} g_1 :=r_1 \chi_{A_1}\\ & \vdots \\ A_k:=\{x\in X : g_{k-1}(x) + r_k \leq f(x) \} \hspace{3ex} & \text{and} \hspace{3ex} g_k:=g_{k-1} + r_k \chi_{A_k} \end{split} \end{equation*} I did a proof that:$\{A_k\}_{k\in { \mathbb{N}}}\in \mathcal{M}$ and i proved until moment that $\lim_{n\to \infty}\sum_{k=1}^{n} r_{k}\chi _{A_{k}}(x) \leq f(x)$ like that: (Just i considered the case where $x\in A_k$ only for finites $k$ and i dont care the case where $x\in A_k$ for all $k$). Because all element of the sequence $B_n=\sum_{k}^{n}{r_k \chi_{A_{k}}}\in \overline {\mathbb{R}}$ (real number extend) satisfy that $B_n(x)\leq f(x)$ for all $x\in X$ then $\lim\sup B_n(x)\leq f(x)$ . I need to show the other inequality for show the equality.

Answer 1

Pick some $x$. Let $I= \{ k | x \in A_k \}$. Note that we always have $\sigma = \sum_{k \in I} r_k \le f(x)$.

If $f(x) = \infty$ then $x \in A_k $ for all $k$ and hence $g_n(x) =r_1+\cdots + r_n$ for all $k$. Hence $\lim_n g_n(x) = \infty$.

Otherwise $f(x) < \infty$. Suppose $\sigma < f(x)$. Let $n$ be the first $n \notin I$ such that $\sigma+r_n \le f(x)$, however, this implies that $x \in A_n$ and so $n \in I$ which is a contradiction. Hence $\sigma = f(x)$.