Piecewise Continuously Differentiable Function/Definition 2 is Continuous

Theorem
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$.

Let $f$ satisfy Piecewise Continuously Differentiable Function/Definition 2.

Then $f$ is continuous.

Proof
$f$ satisfies Piecewise Continuously Differentiable Function/Definition 2.

Therefore, there exists a finite subdivision $\left\{{x_0, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_n = b$, such that $f$ is continuously differentiable on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Here, the derivatives at $x_{i - 1}$ and $x_i$ are understood as one-sided derivatives.

$f$ is differentiable on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, $f$ is continuous at every point of $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$ by Differentiable Function is Continuous.

Here, the continuities at $x_{i - 1}$ and $x_i$ are one-sided.

We use this result to go through every point of every $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ in order to establish the continuity of $f$ there:

$f$ is continuous on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

$\left({x_{i - 1} \,.\,.\, x_i}\right)$ is a subset of $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, $f$ is continuous at every point of $\left({x_{i - 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$.

$x_0$ (= $a$) is the leftmost point in the domain of $f$.

$f$ is right-continuous $x_0$.

Therefore, $f$ is continuous at $x_0$.

$x_n$ (= $b$) is the rightmost point in the domain of $f$.

$f$ is left-continuous $x_n$.

Therefore, $f$ is continuous at $x_n$.

$f$ is continuous at every point $x_i$, $i \in \left\{{1, \ldots, n - 1}\right\}$, because $f$ is both left- and right-continuous at these points.

All in all, we have found that $f$ is continuous at every point of $\left[{x_{i - 1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

$f$ is therefore continuous at every point of the union of those intervals: $\displaystyle \bigcup_{i \mathop = 1}^n \left[{x_{i - 1} \,.\,.\, x_i}\right]$.

$\displaystyle \bigcup_{i \mathop = 1}^n \left[{x_{i - 1} \,.\,.\, x_i}\right]$ equals $\left[{a \,.\,.\, b}\right]$, the domain of $f$.

Therefore, $f$ is continuous as $f$ is continuous at every point of its domain.