Definite Integral of Step Function

Theorem

Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$.

Let $\map f x$ be a step function defined on the interval $\closedint \alpha \beta$:

$\map f x = \lambda_1 \chi_{\mathbb I_1} + \lambda_2 \chi_{\mathbb I_2} + \cdots + \lambda_n \chi_{\mathbb I_n}$

where:

$\lambda_1, \lambda_2, \ldots, \lambda_n$ are real constants

$\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ are intervals, where these intervals partition $\closedint \alpha \beta$

$\chi_{\mathbb I_1}, \chi_{\mathbb I_2}, \ldots, \chi_{\mathbb I_n}$ are characteristic functions of $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$.

Then the definite integral of $f$ {with respect to $x$ over $\closedint \alpha \beta$ is given by:

$\ds \int_\alpha^\beta \map f x \rd x = \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$

where $\alpha_k, \beta_k$ are the endpoints of $\mathbb I_k$ for $1 \le k \le n$.

Each of the intervals $\mathbb I_k$ is such that $f \sqbrk {\mathbb I_k}$ is a constant function:

$\forall x \in \mathbb I_k: \map f x = \lambda_k$

Thus:

$\ds \int_{\mathbb I_k} \map f x \rd x$

$=$

$\ds \int_{\alpha_k}^{\beta_k} \lambda_k \rd x$

$\ds $

$=$

$\ds \lambda_k \paren {\beta_k - \alpha_k}$

$\ds \int_\alpha^\beta \map f x \rd x$

$=$

$\ds \sum_{k \mathop = 1}^n \int_{\mathbb I_k} \map f x \rd x$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$

$\blacksquare$