Question: Let $E$ be a finite measure space and let $f$ be a nonnegative function on $E$. Prove that $f$ is Lebesgue integrable on $E$ if and only if $\sum_{n=0}^\infty 2^nm(\{x\in E:f(x)\geq2^n\})<\infty$.
My Thoughts and Attempt: For the forward direction, if $f$ is integrable on $E$, then $\int_Ef(x)<\infty$. Consider two sets $A=m(\{x\in E:f(x)\geq2^n\})$ and $B=m(\{x\in E:f(x)<2^n\})$. Then, $\int_Ef=\int_Af+\int_Bf$. Since $\int_Ef<\infty$, then, in particular, $\int_Af<\infty$, and so by Markov we get: $2^nm(A)\leq2^n\frac{1}{2^n}\int f(x)<\infty$
For the other direction, suppose $\sum_{n=0}^\infty 2^nm(\{x\in E:f(x)\geq2^n\})<\infty$. So, $\int_Ef=\int_Af+\int_Bf$, and by our hypothesis, we know $\int_Af<\infty$. So, our only concern is about $\int_Bf$….. and this is where I am getting stuck… Any help, suggestions, etc. are greatly appreciated. Also, please let me know if I have any errors in the forward direction. Thank you!
Best Answer
You need $f$ to be measurable.
Let $C_n=\{x:f(x)\ge 2^n\}$ and consider the indicator function $I_{C_n}$. Define $$g(x)=\sum_{n=0}^\infty 2^nI_{C_n}(x).$$ Then $$\int_E g=\sum_{n=0}^\infty 2^nm(C_n).$$ So you want to show $\int_Ef<\infty$ iff $\int_E g<\infty$.
If $2^N\le f(x)<2^{N+1}$ then $g(x)=1+2+4+\cdots+2^N=2^{N+1}-1$. If $f(x)<1$ then $g(x)=0$. Therefore $f(x)\le g(x)+1$ and $f(x)\ge g(x)/2$. These inequalities show that if $\int_Eg$ is finite then so is $\int_Ef$ and vice versa.