Condition for integrability on finite measure space

Question

Let $(X,\mathcal{F},\mu)$ be a finite measure space. If $f:X\rightarrow \mathbb{R}$ is a measurable real function, show that, $f\in L^1(\mu)$ iff $\sum\limits_{n=0}^{\infty}\mu(\{f\geq n\})<\infty$. Am a bit stuck on the $(\Leftarrow)$ direction so any help is appreciated, thanks.

Answer 1

Consider the sets $A_n=\{|f|\geqslant n\}$ and integrate the double inequality $$ \sum_{n=1}^\infty\mathbf 1_{A_n}\leqslant |f|\leqslant\sum_{n=0}^\infty\mathbf 1_{A_n}. $$ Thus, $f$ is integrable if and only if the series $\sum\limits_n\mu(A_n)$ converges.

The hypothesis that $\mu$ is finite is there to ensure that the $n=0$ term $\mu(A_0)=\mu(X)$ is finite.

The double inequality above stems from the pointwise relations, valid for every nonnegative $t$, $$ \lfloor t\rfloor=\sum_{n=1}^\infty\mathbf 1_{t\geqslant n}\leqslant t\lt1+\lfloor t\rfloor=\sum_{n=0}^\infty\mathbf 1_{t\geqslant n}. $$

Answer 2

In general, the $\Leftarrow$ direction does not hold because if we can find $g\geq1$ with $\int g=\infty$. Put $f=-g$, then $\sum_{n=0}^{\infty}\mu([f\geq n])=0$ but $f$ is not integrable.

To avoid typing the sign of absolute value, I modify $f$ to ''$f:X\rightarrow[0,\infty)$''.

For each $n=0,1,2,\ldots,$let $A_{n}=\{x\mid f(x)\in[n,n+1)\}$. Let $B_{n}=[f\geq n]$. Observe that $A_{n}=B_{n}\setminus B_{n+1}$, so $\mu(A_{n})=\mu(B_{n})-\mu(B_{n+1})$. Also, $A_{0},A_{1},\ldots$ are pairwisely disjoint with $\cup_{n}A_{n}=X$.

Suppose that $\int f<\infty$. By Monotone Convergence Theorem, $$\int f=\sum_{n=0}^{\infty}\int_{A_{n}}f\geq\sum_{n}n\mu(A_{n})=\lim_{N\rightarrow\infty}\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)\geq\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)$$ for any $N$. Observe that $\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)=\sum_{n=1}^{N}\mu(B_{n})-N\mu(B_{N+1})$. Therefore, $\sum_{n=0}^{N}\mu(B_{n})\leq N\mu(B_{N+1})+\int f+\mu(B_{0})$. Note that $N\mu(B_{N+1})\leq\int_{B_{N+1}}f\leq\int f.$ Hence, $\sum_{n=0}^{N}\mu(B_{n})\leq2\int f+\mu(B_{0})$. Letting $N\rightarrow\infty$, we have $\sum_{n=0}^{\infty}\mu(B_{n})\leq2\int f+\mu(B_{0})<\infty$.

Conversely, suppose that $\sum_{n=0}^{\infty}\mu(B_{n})<\infty$. Note that $\int f=\sum_{n=0}^{\infty}\int_{A_{n}}f\leq\sum_{n=0}^{\infty}(n+1)\mu(A_{n})$. For each $N$, $\sum_{n=0}^{N}(n+1)\mu(A_{n})=\sum_{n=0}^{N}(n+1)\left(\mu(B_{n})-\mu(B_{n+1})\right)=\sum_{n=0}^{N}\mu(B_{n})-(N+1)\mu(B_{N+1})\leq\sum_{n=0}^{N}\mu(B_{n})$. Letting $N\rightarrow\infty$, we have $\int f\leq\sum_{n=0}^{\infty}(n+1)\mu(A_{n})\leq\sum_{n=0}^{\infty}\mu(B_{n})<\infty$.

Answer 3

This is a lovely application of Borel-Cantelli lemma: by assumption, $\Sigma_n \mu(|f| \geq n) < \infty$, hence the aforementioned lemma tells us $\limsup_n \{|f| \geq n\}$ is a null set.

In other words $$ \mu(\forall n \in \mathbb N\,\exists m \geq n\; |f| \geq m) = 0 $$ which is a convoluted way to say that $f$ is essentially bounded. Therefore $$ \int |f| \,d\mu \leq M\,\mu(X) $$ for some finite constant $M \geq 0$, and $f$ is integrable.