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I have just recently began studying Lebesgue integration theory. I am working through the Stein & Shakarchi Real Analysis: Measure Theory, Integration, & Hilbert Spaces textbook. A problem arises in which the following assertion is made:

If $f$ is a non-negative, measurable, integrable function and $F_k = \{ x \colon 2^k < f(x) \leq 2^{k+1} \}$, then $g(x) \leq f(x) \leq h(x)$ where $$ g(x) = \sum_{k = -\infty}^{\infty} 2^k \chi_{F_k}(x) $$ $$ h(x) = \sum_{k = -\infty}^{\infty} 2^{k+1} \chi_{F_k}(x) $$

I am attempting to prove this statement but I cannot seem to get anywhere. Here is my attempt:

Let $m(\cdot)$ denote the Lebesgue measure. From a previous problem, I have proven that if $f$ is a non-negative integrable function, $\alpha > 0$, and $E_\alpha = \{ x \colon f(x) > \alpha\}$ then $m(E_\alpha) \leq \frac{1}{\alpha} \int f(x) \text{d}x $. Defining $\tilde{F}_k = \{x \colon f(x) > 2^k\}$ and $F_k = \tilde{F}_{k+1} \backslash \tilde{F}_k$, I have $$ m(\tilde{F}_k) \leq 2^{-k} \int f(x) \text{d}x $$ $$ m(\tilde{F}_{k+1}) \leq 2^{-k-1} \int f(x) \text{d}x $$ and thus $$ m(F_k) = m(\tilde{F}_{k+1}) - m(\tilde{F}_k) \leq 2^{-k-1} \int f(x) \text{d}x - 2^{-k} \int f(x) \text{d}x $$ which is to say $$ 2^k m(F_k) = \frac{1}{2}\int f(x) \text{d}x - \int f(x) \text{d}x $$ But this is complete nonsense since measure cannot be negative. Had I not ran into this issue, I was hoping that I could take the infinite sum $\sum_{k = -\infty}^{\infty}(\cdot)$ of both sides, the left side resulting in the expression $\sum_{k = -\infty}^{\infty} 2^k m(F_k) = \int g(x) \text{d}x$. If this had all worked out, getting the $f(x) \leq h(x)$ result would be as simple as multiplying on another factor of $2$ (a rough description of the "proof" but in essence I think correct).

My approach certainly did not work, how else should I go about proving this assertion? Am I making a mistake using the results of previous exercises?

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    $\begingroup$ Hint: note that this statement has nothing to do with measurability of functions or measures of sets, even less with integrability or integrals. $\endgroup$ Oct 19, 2022 at 18:28

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The $F_k$'s form a partition of $f$'s domain (whatever it is) and $$\forall k\in\mathbb Z\quad\forall x\in F_k\quad g(x)=2^k<f(x)\le2^{k+1}=h(x).$$

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    $\begingroup$ Thank you, it all seems so simple now! $\endgroup$ Oct 19, 2022 at 19:45
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    $\begingroup$ moral of the story: when in doubt, draw a picture. $\endgroup$ Oct 19, 2022 at 20:21

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