Let $f:[0,1]\rightarrow \mathbb{R}$ be a measureable function.
Show there exists an $\omega:\mathbb{R} \rightarrow\mathbb{R}$ such that $$\lim_{t \rightarrow \infty}\omega(t)=\infty$$ and $g(t)=\omega(f(t)) $ is integrable on $[0,1]$
Hint: consider the sets $ A_k =\left \{ x: k\leq f(x) < k+1 \right \}$
Any hints or directions? I'm at a loss, the limit condition seems to make it difficult.
Also, I wonder if there's such an omega that is approaching minus infinity at minus infinity.
I'll assume $f:[0,1]\to [0,\infty).$ Because $\sum \mu (A_k) <\infty,$ there exist positive $b_k$ such that $b_k \to \infty$ and $\sum b_k\mu (A_k)<\infty.$ Define $\omega (t) = b_k$ for $t\in [k,k+1).$ Then $\omega \to \infty$ at $\infty.$ We have $\omega \circ f = \sum b_k\chi_{A_k}.$ Thus $\omega \circ f$ is measurable, and
$$\int_0^1 \omega \circ f \, d\mu = \sum b_k\mu (A_k)<\infty.$$