This is a measure theory qualifying exam question: Suppose $f$ is a bounded nonnegative function on $(X,\mu)$ with $\mu(X)=\infty$. Show that $f$ is integrable if and only if$$\sum_{n=0}^\infty\frac{1}{2^n}\mu\left(\left\{x\in X:f(x)>\frac{1}{2^n}\right\}\right)<\infty$$
This is the question exactly as written on the exam, so there is no more information. I was looking through here and have seen many similar questions, but nearly all of them involved a finite measure space/finite measure. I am speculating that maybe the bounded property on $f$ will "take place" of the finite measure property.
In my attempts, if we assume that $f$ is integrable and define $E_n:=\{x\in X:f(x)>\frac{1}{2^n}\}$, then I believe\begin{align*} \int_0^\infty\sum_{n=0}^\infty\frac{1}{2^n}\mu(E_n)dx &= \sum_{n=0}^\infty\frac{1}{2^n}\int_0^\infty\mu(E_n)dx \\ &= \sum_{n=0}^\infty\frac{1}{2^n}\int_Xf(x)d\mu \\ &= ||f||_{L^1(X)}\sum_{n=0}^\infty\frac{1}{2^n}<\infty\end{align*}We only have that this sum is finite for $\mu$-a.e. $x\in [0,\infty)$ trying the problem this way. I don't believe this is the correct way to go about the problem, but I thought I should share what I have tried so far. Any tips for solving this problem are greatly appreciated, thank you.