Prove integrals of step functions are additive wrt interval of integration

For a step function $s(x)$ prove that

$\int_a^c s(x) \, dx + \int_c^b s(x) \, dx = \int_a^b s(x) \, dx \qquad \text{if } a < c < b.$

Proof. Let $P_1 = \{ x_0, \ldots, x_m \}$ be a partition of $[a,c]$ and $P_2 = \{ y_0, \ldots, y_n \}$ be a partition of $[c,b]$ such that $s(x)$ is constant on the open subintervals of $P_1$ and $P_2$ . Then, let $P = P_1 \cup P_2$ so $P$ is a partition of $[a,b]$ and $s(x)$ is constant on the open subintervals of $P$ . Then, let $y_0 = x_m, y_1 = x_{m+1}, \ldots, y_n = x_{m+n}$ . Then,

$\begin{align*} \int_a^c s(x) \, dx + \int_c^b s(x) \, dx &= \sum_{k=1}^m s_k (x_k - x_{k-1}) + \sum_{k=1}^n s_k (y_k - y_{k-1}) \\ &= \sum_{k=1}^m s_k (x_k - x_{k-1}) + \sum_{k=m+1}^{m+n} s_k (x_k - x_{k-1}) \\ &= \sum_{k=1}^{m+n} s_k (x_k - x_{k-1}) \\ &= \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}$

Stumbling Robot

Prove integrals of step functions are additive wrt interval of integration

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply