Integrability on Finite Measure Spaces via Tail Integral

Theorem 1 (Integrability via Tail Integral on Finite Measure Spaces).

Let $(X,\mathcal{M},\mu )$ be a finite measure space, meaning $\mu (X)<\infty$. A measurable function $f:X\to \overline{\mathbb{R}}$ or $f:X\to \mathbb{C}$ is integrable if and only if ${\lim }_{t\to \infty }{\int }_{\{|f|\ge t\}}|f|d\mu =0$.

Proof. (⇒) Assume $f$ is integrable, which means ${\int }_X|f|d\mu <\infty$. Let $g_t(x)=|f(x)|{\chi }_{\{|f|\ge t\}}(x)$, where $\chi$ is the characteristic function. For each $x\in X$ such that $|f(x)|<\infty$, ${\lim }_{t\to \infty }{g}_t(x)=0$. Since $f$ is integrable, $|f(x)|<\infty$ for almost every $x$. Thus, $g_t\to 0$ almost everywhere. The function $|f|$ is integrable, and $|g_t(x)|\le |f(x)|$ for all $t>0$ and $x\in X$. By the Dominated Convergence Theorem:

$$\underset{t\to \infty }{\lim }{\int }_{\{|f|\ge t\}}|f|d\mu =\underset{t\to \infty }{\lim }{\int }_X{g}_{t}d\mu ={\int }_X\underset{t\to \infty }{\lim }{g}_{t}d\mu ={\int }_{X}0d\mu =0.$$

(⇐) Assume ${\lim }_{t\to \infty }{\int }_{\{|f|\ge t\}}|f|d\mu =0$. To show $f$ is integrable, we must show ${\int }_X|f|d\mu <\infty$. For any constant $M>0$, the integral is decomposed as:

$${\int }_X|f|d\mu ={\int }_{\{|f|<M\}}|f|d\mu +{\int }_{\{|f|\ge M\}}|f|d\mu .$$

The first term on the right is bounded:

$${\int }_{\{|f|<M\}}|f|d\mu \le {\int }_{\{|f|<M\}}Md\mu =M\mu (\{|f|<M\}).$$

Since $\mu (X)<\infty$, we have $\mu (\{|f|<M\})\le \mu (X)<\infty$. Thus, ${\int }_{\{|f|<M\}}|f|d\mu$ is finite. The second term, ${\int }_{\{|f|\ge M\}}|f|d\mu$, approaches $0$ as $M\to \infty$ by hypothesis, and so is finite for any sufficiently large $M$. The integral ${\int }_X|f|d\mu$ is the sum of two finite quantities, hence it is finite. $\square$