Heaviside Step Function is Piecewise Continuous

Theorem
Let $c \ge 0$ be a constant real number

The Heaviside step function:


 * $\mu_c\left({t}\right) = \begin{cases}

1 & : t > c \\ 0 & : t < c \end{cases}$

is piecewise continuous for any interval of the form:


 * $\left[{c - M \,.\,.\, c + M}\right]$

where $M > 0$ is some arbitrarily large constant.

Proof
Let the finite subdivision of $\mu_c$ be:


 * $\left\{{c - M, c, c + M}\right\}$

Let $\epsilon > 0$.

For $t \to \left({c - M}\right)^+$, choose $\delta$ to be any number at all, because:


 * $c - M < t < c - M + \delta \implies \left\vert{\mu_c\left({t}\right) - 0}\right\vert = \left \vert { 0 - 0 }\right\vert < \epsilon$

holds, from True Statement is implied by Every Statement.

The case for $t \to c^-$ is proved similarly.

For $t \to c^+$, again, choose $\delta$ to be any number at all, because:


 * $c - \delta < t < c \implies \left\vert{\mu_c\left({t}\right) - 1}\right\vert = \left \vert { 1 - 1 }\right\vert < \epsilon$

holds, from True Statement is implied by Every Statement.

The case for $t \to \left({c + M}\right)^-$ is proved similarly.

For $t < c$, then $\mu_c\left({t}\right) = 0t + 0$, and the result holds from Linear Function is Continuous.

The $t > c$, then $\mu_c\left({t}\right) = 0t + 1$, and the result again holds from Linear Function is Continuous.