Prove the additive property of integrals of step functions

For $s(x), t(x)$ step functions defined on an interval $[a,b]$ , prove

$\int_a^b (s(x) + t(x)) \, dx = \int_a^b s(x) \, dx + \int_a^b t(x) \, dx.$

Proof. Let $P = \{ x_0, x_1, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x)+t(x)$ is constant on the open subintervals of $P$ (we know such a $P$ exists since we can take the common refinement of the partition of $[a,b]$ on which $s(x)$ and $t(x)$ individually are constant on open subintervals). Let $s(x) + t(x) = s_k + t_k$ if $x_{k-1} < x < x_k$ for $k=1,2, \ldots, n$ . Then, working from the definition of the integral of a step function on an interval, we have

$\begin{align*} \int_a^b (s(x)+t(x))\, dx &= \sum_{k=1}^n (s_k + t_k)(x_k - x_{k-1}) \\ &= \sum_{k=1}^n \left( s_k (x_k - x_{k-1}) + t_k (x_k - x_{k-1}) \right) \\ &= \sum_{k=1}^n s_k (x_k - x_{k-1}) + \sum_{k=1}^n t_k (x_k - x_{k-1}) \\ &= \int_a^b s(x) \, dx + \int_a^b t(x) \, dx. \qquad \blacksquare \end{align*}$

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Prove the additive property of integrals of step functions

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