Prove the comparison theorem for integrals of step functions

For step functions $s(x) < t(x)$ for every $x \in [a,b]$ prove that

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

Proof. Let $P = \{ x_0, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x)$ and $t(x)$ are constant on the open subintervals of $P$ (such a partition exists since we can take the common refinement of partitions on which $s$ and $t$ are constant on open subintervals). Then, assume $s(x) = s_k, \ t(x) = t_k$ if $x_{k-1} < x < x_k$ . From the definition of integrals of step functions, we have

$\int_a^b s(x) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}) \qquad \text{and} \qquad \int_a^b t(x) \, dx = \sum_{k=1}^n t_k (x_k - x_{k-1}).$

Then, $s_k < t_k$ (since $s(x) < t(x)$ for every $x \in [a,b]$ ), which implies,

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

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Prove the comparison theorem for integrals of step functions

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