Example

For every positive measure $\mu$ on $\mathbf{R}$, every step function with compact support is $\mu$-integrable; for, the characteristic function of an open (resp. closed) interval is lower (resp. upper) semi-continuous, and every step function is a linear combination of such characteristic functions. It follows that if $\mathbf{f}$ is a regulated function on $\mathbf{R}$ with compact support (FRV, II, §1, No. 3), then $\mathbf{f}$ is integrable, because it is the uniform limit of a sequence of step functions $\mathbf{g}_n$ with support contained in a fixed compact set (No. 3, Prop. 3); moreover, $\int \mathbf{f}\, d\mu = \lim_{n \to \infty} \int \mathbf{g}_n\, d\mu$.

If, in particular, $\mu$ is taken to be Lebesgue measure, one sees that for every regulated function $\mathbf{f}$ with compact support, the integral $\int \mathbf{f}\, d\mu$ is equal to the integral $\int_{-\infty}^{+\infty} \mathbf{f}(x)\, dx$ defined in FRV, II, §2, No. 1.