Prove that $f\in L^1(A)\Leftrightarrow \sum_{n}^{\infty}m(\{ x\in A : f(x)\geq n \}) < \infty$

I'm stuck with some problem of my Integral Calculation in Several Variables course. The problem goes like this:

Let $A\subset \mathbb{R}$ be a measurable set with $m(A)<\infty$, and $f:A\longrightarrow [0,\infty)$ a Lebesgue-measurable function. Prove that: $$f\in L^1(A)\Longleftrightarrow \sum_{n}^{\infty}m(\{ x\in A : f(x)\geq n \}) < \infty.$$

The notation I used is:

• $m$ as the Lebesgue measure function
• $L^1(A)=\{ f:A\rightarrow \mathbb{\overline{R}} : \int_{A}|f|\,\mathrm{d}m<+\infty \}$

I've started defining the set $A=f^{-1}([0,\infty))$ as a numerable sum of disjoint measurable sets (because it's said it's measurable) $\sum^{\infty}_{k=0}\cup I_k$, being each $I_k$ the real interval $[k,k+1)$. I imagine I should come to some conclusion like that any unbounded (upper bound) sets of $f(x)$ with $(x\in A)$ have measure $0$.

Solution 1:

As stated in the comment you have that $$ f^{-1}([n,\infty ))= \bigcup_{k\geqslant n}f^{-1}([k,k+1)) $$ Therefore $$ \begin{align*} \sum_{n\geqslant 0}m(f^{-1}([n,\infty ))&= \sum_{n\geqslant 0}m\left( \bigcup_{k\geqslant n}f^{-1}([k,k+1))\right)\\ &= \sum_{n\geqslant 0}\sum_{k\geqslant n}m(f^{-1}([k,k+1))\\ &\overset{(*)}{=} \sum_{k\geqslant 0}\sum_{0\leqslant n\leqslant k}m(f^{-1}([k,k+1))\\ &=\sum_{k\geqslant 0}(k+1)m(f^{-1}([k,k+1))\\ &= \sum_{k\geqslant 0}\int(k+1)\mathbf{1}_{f^{-1}([k,k+1)} \mathop{}\!dm\\ &\overset{(**)}{=} \int\sum_{k\geqslant 0}(k+1)\mathbf{1}_{f^{-1}([k,k+1)} \mathop{}\!dm\\ &\geqslant \int f\mathop{}\!dm \end{align*} $$ where we used Tonelli's (or the monotone convergence) theorem in $(*)$ and $(**)$. Similarly $$ \begin{align*} \int f \mathop{}\!dm &\geqslant \int\sum_{k\geqslant 0}k \,\mathbf{1}_{f^{-1}([k,k+1)}\mathop{}\!dm\\ &=\sum_{k\geqslant 0}k\, m(f^{-1}([k,k+1))\\ &=\sum_{k\geqslant 1}\sum_{1\leqslant n\leqslant k} m(f^{-1}([k,k+1))\\ &= \sum_{n\geqslant 1} \sum_{k\geqslant n}m(f^{-1}([k,k+1))\\ &= \sum_{n\geqslant 1} m(f^{-1}([n,\infty )) \end{align*} $$ Finally note that $m(f^{-1}[0,\infty ))= m(A)<\infty $, so the answer is clear.