Step Function satisfies Dirichlet Conditions

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 $\openint \alpha \beta$.

Then $f$ satisfies the Dirichlet conditions.

Proof
Recall the definition of step function:

Recall the Dirichlet conditions:

We inspect the Dirichlet conditions in turn.

$(\text D 1)$: Absolute Integrability
From Definite Integral of Step Function:
 * $\ds \int_\alpha^\beta \map f x \rd x = \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$

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

When $\lambda_k < 1$, we have that:
 * $\size {\lambda_k} = -\lambda_k$

and so:
 * $\ds \int_{\alpha_k}^{\beta_k} \lambda_k \rd x = -\lambda_k \paren {\beta_k - \alpha_k}$

Absolute integrability follows.

$(\text D 2)$: Finite Number of Local Maxima and Minima
By nature of $f$ being a step function, the image set of $f$ is finite:


 * $f \openint \alpha \beta = \set {\lambda_1, \lambda_2, \ldots, \lambda n}$

The set of local maxima and the set of local minima must be subsets of $f \openint \alpha \beta$.

Therefore there must be a finite number of each.

$(\text D 3)$: Discontinuities are Finite in Number and Nature
From Constant Real Function is Continuous, there are no discontinuities except perhaps at the endpoints of $\mathbb I_k$ for $1 \le k \le n$.

Let us inspect those endpoints.

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

Let $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ be ordered such that:
 * $\forall k: \alpha_k < \alpha_{k + 1}$

By definition, $\beta_{k - 1} = \alpha_k$.

Thus we have that:

and if $\lambda_k \ne \lambda_{k - 1}$ then this is indeed a finite discontinuity.

By the nature of the step function, there are a finite number of them.

Thus all Dirichlet conditions are satisfied by $f$.