Show that $f = \sum_{k=1}^{\infty} \frac{1}{k} \mathbb{1}_{A_k}$ is a representation for a measurable function I want to show that for a (Borel-)measurable Function $f \in \mathcal{M}^+(\Omega,\mathfrak{S})$, $f: \Omega \rightarrow [0,\infty])$ exists a representation of the form
$$f = \sum_{k=1}^{\infty} \frac{1}{k} \mathbb{1}_{F_k}$$
for $\mathfrak{S}$ measurable $F_k$.
I think I need to look at $F_1:=[f\geq 1]$ and for $k\geq2$:
$$F_k:=[f\geq \frac{1}{k} + \sum_{i=1}^{k-1} \frac{1}{i} \mathbb{1}_{F_i}]$$
but I have no idea if that representation even makes sense.
Any help/solutions would be greatly appreciated.
 A: OP's construction indeed works.
1. Let $f_0 = 0$, and we define $(f_k)_{k\geq 1}$ and $(F_k)_{k\geq 1}$ recursively as follows: Suppose $ f_{k-1}$ has been defined as a measurable function satisfying $f_{k-1} \leq f$ on $\Omega$. Then let
$$ F_k = \{\omega \in \Omega : f(\omega) \geq \frac{1}{k} + f_{k-1}(\omega)\}. $$
Since both $f$ and $f_k$ are measurable, $F_k$ is also a measurable set. Then let
$$f_k = f_{k-1} + \frac{1}{k} \mathbf{1}_{F_k}.$$
Since both $f_{k-1}$ and $F_k$ are measurable, so is $f_k$. Moreover, the induction hypothesis and the definition of $F_k$ together show that $f_k \leq f$ on $\Omega$.
2. By the construction, $(f_k)$ is an increasing sequence of measurable functions, and so, the function $\tilde{f}$ defined by
$$ \tilde{f} = \lim_{k\to\infty} f_k = \sum_{k=1}^{\infty} \frac{1}{k} \mathbf{1}_{F_k}$$
is also a measurable function that still satisfies $\tilde{f} \leq f$ on $\Omega$. Now we show that $f = \tilde{f}$. Assume otherwise that there exists $\omega \in \Omega$ such that $\tilde{f}(\omega) < f(\omega)$. In particular, $\tilde{f}(\omega)$ is finite. Let $\varepsilon = f(\omega) - \tilde{f}(\omega) > 0$. Then $\omega \in F_k$ holds for any $k$ satisfying $\frac{1}{k} \leq \varepsilon$, and so,
$$ \tilde{f}(\omega) \geq \sum_{k \geq 1/\varepsilon} \frac{1}{k} = \infty, $$
a contradiction! Therefore $\tilde{f} = f$.
