# Prove $g(x) \leq f(x) \leq h(x)$ for all $x$ for specifically defined $g(x)$ and $h(x)$

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?

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

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