$\int f\,d\mu<\infty\Leftrightarrow\sum_{n=0}^\infty\frac{1}{2^n}\mu\left(\left\{x\in X:f(x)\geq\frac{1}{2^n}\right\}\right)<\infty$

Question

I have to do a measurement theory exercise but I am stuck and I hope someone can help me.

The statement is as follows: Let $f\in M​​^+(X,S)$ be a measurable, positive and bounded function. Show that $\int f\,d\mu<\infty\Leftrightarrow\sum_{n=0}^\infty\frac{1}{2^n}\mu\left(\left\{x\in X:f(x)\geq\frac{1}{2^n}\right\}\right)<\infty$.

I can assume that $\mu(X)=\infty$.

Here is what I have so far:

$A_n:=\left\{x\in X:f(x)\geq\frac{1}{2^n}\right\}$

($\Rightarrow$)

$\frac{\mu(A_n)}{2^n}\leq\int_{A_n}f\,d\mu$

$\sum_{n=0}^{\infty}\frac{\mu(A_n)}{2^n}\leq\sum_{n=0}^{\infty}\int_{A_n}f\,d\mu=\sum_{n=0}^{\infty}\int f\cdot\mathcal{X}_{A_n}\,d\mu=\int\sum_{n=0}^{\infty}f\cdot\mathcal{X}_{A_n}\,d\mu$

Next, I would like to say that $\int\sum_{n = 0}^{\infty}f\cdot\mathcal{X}_{A_n}\,d\mu<\infty$, but I don't know how to justify.

($\Leftarrow$)

$\int f\,d\mu=\sum_{n=0}^\infty\int_{A_n-A_{n-1}}f\,d\mu\leq\sum_{n=1}^{\infty}\frac{\mu(A_n-A_{n-1})}{2^{n-1}}\leq\sum_{n=1}^{\infty}\frac{\mu(A_n)}{2^{n-1}}<\infty$

If anyone can check out what I did and help me with the first part of the demo, I would really appreciate it.

Answer 1

As in the second part you get the following for the first part: $\int f\,d\mu=\sum_{n=0}^\infty\int_{A_n-A_{n-1}}f\,d\mu\geq\sum_{n=1}^{\infty}\frac{\mu(A_n-A_{n-1})}{2^{n}}$. Does that tell you that $\sum \frac {\mu (A_n)} {2^{n}} <\infty$?