Remark 1

Let $\mathbf{f}$ be a regulated function on $\mathbf{R}$ that is integrable with respect to Lebesgue measure $\mu$; then $|\mathbf{f}|$ is also integrable (No. 2, Cor. 1 of Prop. 1), and, setting $I_n = [-n, n]$, $|\mathbf{f}|$ is the upper envelope of the increasing sequence of regulated functions $|\mathbf{f}| \varphi_{I_n}$, therefore $\int |\mathbf{f}|\, d\mu = \lim_{n \to \infty} \int_{-n}^{n} |\mathbf{f}(x)|\, dx$ by Th. 2 of No. 3; thus, the integral $\int_{-\infty}^{+\infty} f(x) dx$ is absolutely convergent (FRV, II, §2, No. 3). Moreover, $\int f\, d\mu = \int_{-\infty}^{+\infty} f(x) dx$ by Th. 2 of No. 3. Conversely, suppose that $\int_{-\infty}^{+\infty} f(x) dx$ is absolutely convergent; again, by Th. 2 of No. 3, $\int f\, d\mu = \int_{-\infty}^{+\infty} f(x) dx$. Note that if the integral $\int_{-\infty}^{+\infty} f(x) dx$ is convergent but not absolutely convergent; then $f$ is not integrable with respect to Lebesgue measure.