Let $\mu(A)<\infty$, and the function $f(x)$ is measurable and finite on $A$. Suppose that $E_k(f)=\{x\in A: k\leq |f(x)|<k+1\}$ for $k\geq 0$. Prove that $f(x)\in L(A)$ iff $$\sum \limits_{k=1}^{\infty}k\mu(E_k(f))<\infty.$$
$\Rightarrow$ Suppose that $f(x)\in L(A)$ then $|f(x)|\in L(A)$. Since $f(x)$ is finite on $A$ then $A=\sqcup_{k=0}^{\infty} E_k(f)$. For each $n\geq 0$ consider the function $F_n(x)=\sum \limits_{k=0}^{n}|f(x)|\chi_{E_k}(x)$ then $\{F_n(x)\}_{n=0}^{\infty}$ be a family of measurable functions on $A$ such that $F_n(x)\to |f(x)|$ for each $x\in A$ and $|F_n(x)|\leq |f(x)|$ for all $n\geq 1$. Then by Lebesgue dominated convergence theorem it follows that $$\int \limits_{A}|f(x)|d\mu=\lim \limits_{n\to \infty}\int \limits_{A}F_n(x)d\mu=\sum \limits_{k=0}^{\infty} \int \limits_{A}|f(x)|\chi_{E_k}(x)d\mu\geq $$ $$\sum \limits_{k=0}^{\infty} \int \limits_{A}k\chi_{E_k}(x)d\mu=\sum \limits_{k=0}^{\infty} k\mu(E_k)<\infty.$$
$\Leftarrow$ Suppose that $\sum \limits_{k=0}^{\infty} k\mu(E_k)<\infty$. We see that functions $\{F_n(x)\}_{n=0}^{\infty}$ are nonnegative, non-decreasing measurable functions on $A$ such that $F_n(x)\to |f(x)|$ for each $x\in A$. Then by monotone convergence theorem $$\int\limits_{A}|f(x)|d\mu=\lim \limits_{n\to \infty}\int \limits_{A}F_n(x)d\mu=\sum \limits_{k=0}^{\infty}\int\limits_{A}|f(x)|\chi_{E_k}(x)d\mu<\sum \limits_{k=0}^{\infty}\int\limits_{A}(k+1)\chi_{E_k}(x)d\mu=$$$$=\sum \limits_{k=0}^{\infty}(k+1)\mu(E_k)<\infty$$ because $\sum \limits_{k=0}^{\infty}k\mu(E_k)<\infty$.
I checked my solution and I have never used the fact that $\mu(A)<\infty$. What's wrong with solution?
