The question: Suppose we let the measure space be $[0,1]$, along with the Lebesgue measure on $[0,1]$ on $\lambda$. Assume that $g(x) > 0$ almost everywhere on $[0,1]$ and that $g$ is measurable. Let $C_k = \{ x \in [0,1] \ | g(x) \geq 3^k \}$. Show g is integrable if and only if $\sum_{k = 0}^{\infty} 3^{k} \lambda(C_k) < \infty$.
I am using this as a source to solve this problem: Show that $f \in L^{1}(X)$ if and only if $\sum_{n=1}^{\infty} n \mu(E_{n}) < \infty$.
Here is what I got. I defined $E_k = \{ x \in [0,1] : 3^k \leq g(x) \leq 3^{k+1} \}$. Then $C_k = \bigcup_{j = k}^{\infty} E_j$. Now, I have that
$g(x) = \sum_{k = 0}^{\infty} g 1_{C_k} $. I am not sure how to approach this problem. Can you give some comments on how to solve this problem?
Thank you for your help!
Best Answer
You idea is good ! By assumption : $$[0,1] = \{ 0 < g < 1\} \cup \bigcup_{k \geq 1} E_k$$ Concerning the first implication. Suppose the sum finite. Since g is postive, we can use Fubini (2nd equality) : $$\int g = \int \sum_{k \geq 0} I_{E_k} g = \int_{E_0} g + \sum_{k \geq 1} \int_{E_k} g \leq 1 + 3 \sum_{k \geq 1} 3^k \lambda(C_{k+1}) < \infty$$ For the other implication. Suppose g intégrable. By Markov and Fubini: $$\infty > \int g \geq \int \sum_{k \geq 1} I_{E_k} g \geq \sum_{k \geq 1} \int_{C_k} g \geq \sum_{k \geq 1} 3^k \lambda(C_k)$$ (Markov : $g I_{g \geq 3^k} \geq 3^k I_{g \geq 3^k}$)