Exercise 1.3.18 (Finite and null integrals criteria)

Throughout the proof, we will make use of the infamous measure-theoretic convention "$\infty \times 0=0$".

(i)

We argue by contradiction. Suppose that ${\int }_{{ℝ}^d}f<\infty$ and $f$ is infinite on a non-null set $E$. We then have $${\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f\ge {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f\cdot 1_E={\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\infty \cdot 1_E=\infty \cdot m(E)=\infty$$

since $m(E)\ne 0$. For the converse statement, consider any constant function $f=1$, and we have

$${\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}1\cdot 1_{{ℝ}^d}=1\cdot m({ℝ}^d)=\infty .$$

(ii)

Suppose that ${\int }_{{ℝ}^d}f=0$. By Markov’s inequality (Lemma 1.3.15) and by monotonicity we have for any $\lambda >0$ $$m\left(\left\{x\in {ℝ}^d:f(x)\ne 0\right\}\right)\le m\left(\left\{x\in {ℝ}^d:f(x)\ge \lambda \right\}\right)\le \frac{1}{\lambda }{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f=0$$

Conversely, suppose that $f$ is zero almost everywhere. We then have by divisibility property of the Lebesgue integral

$${\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f={\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f\cdot 1_{\left\{x\in {ℝ}^d:f(x)=0\right\}}+{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f\cdot 1_{\left\{x\in {ℝ}^d:f(x)\ne 0\right\}}\le 0\cdot \infty +\sup <!--\; FUNCTION\; APPLICATION\; -->\{f\}\cdot 0=0.$$