# Heaviside Step Function is Piecewise Continuous

Jump to navigation
Jump to search

## Theorem

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

- $\mu_c \left({t}\right) = \begin{cases} 1 & : t > c \\ 0 & : t < c \end{cases}$

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 all other $t$, we invoke Continuity of Heaviside Step Function.

$\blacksquare$