Suppose $E \subset \mathbb{R}^d$ and $mE < \infty$. Let $f(x)$ be a measurable function defined on $E$, and $0 < f(x) < \infty$ a.e. in $E$. Prove that $\forall 0 < \epsilon < mE$, there exists a subset $G \subset E$, and $k_0 \in \mathbb{N}$, such that $$m(E-G) < \epsilon \quad \text{and} \quad \frac{1}{k_0} \leq f(x) \leq k_0, \quad \forall x \in G$$ Proof Attempt Claim: for any $\epsilon > 0$ there exists a closed set $G$ such that $m(E - G) < \epsilon$. This is true because, since $E$ is measurable $E^c$ is measurable. Since $E^c$ is measurable then, by definition, there exists an $O \supset E^c$ such that $m(O - E^c) < \epsilon$. Take $G$ to be $O^c$. Clearly, $G$ is closed and $G \subset E$. Also, $E-G= O - E^c \implies m(E - G) < \epsilon$.
Now, note that since $mE < \infty$, we can choose $k_0 \in \mathbb{N}$ to be the smallest integer such that $mE < k_0$...
...feel like I'm doing something wrong. Please avoid $\sigma$-algebra advice.
Let $G_n=\{x\in E:\frac 1 n \leq f(x) \leq n\}$. Then $G_n$ increases to $E$ so $m(G_n)$ tends to $m(E)$. Choose $n$ such that $m(E) <m(G_n) -\epsilon$. Then $m(E\setminus G_n) <\epsilon$ and $\frac 1 {k_0} \leq f(x) \leq k_o$ on $G_n$ where $k_0=n$. Take $G=G_n$.