Continuity of Heaviside Step Function

Theorem
Let $\mu_c: \R \to \R$ be the Heaviside step function:


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

0 & : x < c \\ 1 & : x > c \\ \text{arbitrary} & : x = c \end{cases}$

Then $\mu_c$ is continuous at every point of $\R$ except at $c$.

Proof
Let $x \in \R: x \ne c$.

Let $\epsilon \in \R_{>0}$.

Let $\delta < \left\vert{x - c}\right\vert$.

Then by definition of the Heaviside step function:
 * $\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({x}\right) = \begin{cases}

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

Thus:
 * $\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({y}\right) = \mu_c \left({x}\right)$

Thus:
 * $\left\vert{y - x}\right\vert \implies : \mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$

and so $\mu_c$ is continuous at $\R$ except at $x$.

Now suppose $x = c$.

Let $\epsilon = \dfrac 1 3$.

Let $\delta \in \R_{>0}$.

Let $\alpha \in \R_{>0}: \alpha < \delta$.

Let $y = x + \alpha$.

Let $y \in \R: \left\vert{y - x}\right\vert < \delta$ and $y \ne x$.

Suppose $\mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$.

Then:

while:

So whatever the value of $f \left({x}\right)$, either:
 * $f \left({x + \alpha}\right) - f \left({x}\right) > \epsilon$

or:
 * $f \left({x - \alpha}\right) - f \left({x}\right) > \epsilon$

It follows that $\mu_c$ is not continuous at $c$.