For a Lebesgue measurable sets and functions problem, I need to prove this statement:
Being $A\subset\mathbb{R}$ a measurable set with $m(A)<\infty$, and $f:A\rightarrow[0,\infty)$ a Lebesgue measurable function: $$\sum_{n=0}^\infty(m(\{x\in A:f(x)\geq n\}))<\infty\Longleftrightarrow \int_{\mathbb{R}}\sum_{k=0}^\infty (k\cdot\chi_{f^{-1}([k,k+1))}) dm<\infty$$
The notation I'm using is: $m$ for the Lebesgue measure and $\chi_B$ the characteristic function of $B$, to be said, $\chi_B(x)=1\Leftrightarrow x\in B$ and $\chi_B(x)=0\Leftrightarrow x\notin B$.
I've started calling $S=\sum_{k=0}^\infty (k\cdot\chi_{f^{-1}([k,k+1))}$, so that $S$ is in fact a simple function, and that means (by characterisation of simple functions) $$\int_RS=\sum_{k=0}^\infty (k\cdot m(f^{-1}([k,k+1)))$$ I clearly see that $\sum_{k=0}^\infty (m(f^{-1}([k,k+1)))=m(f^{-1}([0,\infty)))=m(A)=<\infty$, because the problem says $m(A)<\infty$. My problem is that I'm struggling with the multiplying $k$ inside of the sum. I don't know what to (guess using $\sum_{n=0}^\infty(m(\{x\in A:f(x)\geq n\}))<\infty$, but I don't know how) do with that $k$ to prove that $$\sum_{k=0}^\infty (k\cdot m(f^{-1}([k,k+1)))<\infty.$$ I will thank any answer or comment.
Hint: Let $a_n=m\{x: f(x) \in [k,k+1)\}$ and $s_n=a_n+a_{n+1}+...=m\{x: f (x) \geq n\}$. Then $\sum k a_k =a_1+2a_2+3a_3+...=s_1+s_2+...=\sum m(f(x) \geq n)$. To finish the proof you have to know that the sum and integral on RHS can be interchanged. But that is true by non-negativity.