Show that $f \in L^{1}(\mu)$ iff $\sum_{n=1}^{\infty}n \mu(B_n) < \infty$ iff $\sum_{n=1}^{\infty}{\mu (A_n)} < \infty$ Let $(X, \mathcal{M}, \mu )$ be a space of finite measure and $f \colon X \rightarrow \mathbb{C}$. For each $n \in \mathbb{N}$, let $A_n = \{ x \in X \colon |f(x)|>n \}$ and $B_n = \{ x \in X \colon n \leq |f(x)|< n+1\}$. Show that the following statements are equivalent:
(a) $f \in L^{1}(\mu)$ 
(b)  $\sum_{n=1}^{\infty}n \mu(B_n) < \infty$
(c) $\sum_{n=1}^{\infty}{\mu (A_n)} < \infty$
Definition: We define $L^1(\mu)$ as the collection of all measurable complex functions $f$ in $X$ for which $\int_{X}{|f|}d \mu < \infty$. The elements of $L^1 (\mu)$ are called Lebesgue integrable functions (with respect to $\mu$) or summable functions.
Definition 2: If $f = u + iv$, where $u$ and $v$ are real functions measurable in $X$, and if $f \in L^1(\mu)$, we define $\int_{E}{f} d\mu = \int_{E}{u^{+}} d\mu - \int_{E}{u^{-}}  d\mu + i\int_{E}{v^{+}} d\mu - i\int_{E}{v^{-}} d\mu $ for each measurable set $E$.
Here $u^{+}$ and $u^{-}$ are the positive and negative parts of $u$; $v^{+}$ and $v^{-}$ are obtained analogously from $v$. These four functions are measurable, real, and non-negative.
I wanna do a test for this exercise, but I really don't have a good idea to do it. To show that (a) implies (c), for example, I have thought about defining the function, $f$ let's say: $f(x) = \sum_{n=1}^{\infty}{\chi_{A_n}}(x)$, where $\chi_{A_n}$ is the characteristic function of $A_n$. Then use the following theorem: If $f_n \colon X \rightarrow [0, \infty]$ is measurable, for $n = 1, 2, 3,. . . ,$ and $f(x) = \sum_{n=1}^{\infty}{f_{n}(x)}$ for all $x \in X$, then $\int_{X}{f}d\mu = \sum_{n=1}^{\infty}{\int_{X}{f_{n}d\mu}}$.
To then obtain that $\infty > \int_{X}f d\mu = \sum_{n=1}^{\infty}{\int_{X}{\chi_{A_n}}d\mu} = \sum_{n=1}^{\infty}{\mu (A_n)}$. I'm not sure of my reasoning because the function $f$ is really a complex function. I need some help to do this.
 A: Here is an attempt. (a) $\Rightarrow$ (b).
$$\sum_{n \in \mathbb{N}}n\mu(B_n)=\sup_{N \in \mathbb{N}}\sum_{n=1}^Nn\mu(B_n)=\sup_{N \in \mathbb{N}}\sum_{n=1}^Nn\int\mathbb{I}_{B_n}(x)\mu(dx)=$$
$$=\sup_{N \in \mathbb{N}}\int \bigg(\sum_{n=1}^Nn\mathbb{I}_{B_n}(x)\bigg)\mu(dx)=\sup_{N \in \mathbb{N}}\int f_N(x)\mu(dx)$$
we have that $f_N \to g\leq|f| \in \mathcal{L}^1(\mu)$ pointwise and $g$ is positive and measurable so $g \in \mathcal{L}^1(\mu)$ and by DCT
$$\sup_{N \in \mathbb{N}}\int f_N(x)\mu(dx)=\int g(x)\mu(dx)<\infty$$
(b) $\Rightarrow$ (c). If $|f(x)|>n$ then $\mathbb{I}_{A_n}(x)=1$ and $n\mathbb{I}_{B_n}(x)\geq n$. If $|f(x)|=n$ then $\mathbb{I}_{A_n}(x)=0$ and  $n\mathbb{I}_{B_n}(x)=n$ . If $|f(x)|<n$ then $\mathbb{I}_{A_n}(x)=0$ and  $n\mathbb{I}_{B_n}(x)=0$. Therefore
$$\mathbb{I}_{A_n}(x)\leq n \mathbb{I}_{B_n}(x)$$
$$\int\mathbb{I}_{A_n}(x)\mu(dx)\leq \int n \mathbb{I}_{B_n}(x)\mu(dx)$$
$$\sum_{n=1}^\infty \mu(A_n)\leq \sum_{n=1}^\infty n\mu(B_n) < \infty$$
(c) $\Leftrightarrow$ (a). Is answered here (as suggested in the comments, use monotone convergence if you do not want to use Tonelli).
A: *

*From
$$\begin{align}
n \mu(n<|f|\leq n+1)&\leq \int_{n<|f|\leq n+1}|f|\,d\mu\leq (n+1)\mu(n\leq |f|\leq n+1)\\
&\leq 2n\mu(n\leq |f|\leq n+1)\end{align}$$
one gets that
$$\sum_n n\mu(n<|f|\leq n+1)\leq \int |f|\,d\mu\leq 2\sum_n n\mu(n<|f|\leq n+1)$$
So convergence of $\int|f|\,d\mu$  and $\sum_nn\mu(n<|f|\leq n+1)$ are equivalent.


*For the other equivalencies, notice that
$$\begin{align}
\sum^\infty_{n=1}\mu(|f|>n)&=\sum^\infty_{n=1}\sum^\infty_{m=n}\mu(m<|f|\leq m+1)=\sum^\infty_{m=1}\sum^m_{n=1}\mu(m<|f|\leq m+1)\\
&=\sum^\infty_{m=1}m\mu(m<|f|\leq m+1)\end{align}$$
The conclusion follows immediately.
