Why is the Indefinite Integral of a Step Function and continuous? As I posted before, say I have:\begin{equation}
  S(X)=\left\{
  \begin{array}{@{}ll@{}}
    c_0, & \text{if}\ t_0 \leq x <t_1 \\
    c_1, & \text{if}\ t_1 \leq x <t_2 \\ 
    c_2, & \text{if}\ t_2 \leq x <t_3 \\ 
. & \ . \  \\ 
. & \ . \  \\
. & \ . \  \\
. & \ . \  \\
c_{n-1}, & \text{if}\ t_{n-1} \leq x \leq t_n \\\end{array}\right.
\end{equation} .  
Why is the indefinite integral of S(x) piecewise linear and continuous?
I think I understand why it it is piecewise linear (hopefully as I was trying to do in my last post) but why is the indefinite integral of a step function necessarily continuous? It is not clear to me why the integral of $S(x)$, or an arbitrary step function for that matter cannot have jump discontinuities.  Thanks for the help.
Here is a picture stating that it must necessarily be continuous:

Perhaps it would help me if someone could help me understand how they calculated the indefinite integral of this step function:

Essentially they have in this picture:
$T(X) =1$ for $0 \leq x < 2$
$T(X) =-1$ for $2 \leq x \leq 4$.
They get the indefinite integral to be:
$x$ for $0 \leq x < 2$
$4-x$ for $2 \leq x \leq 4$.
How did they compute this?
 A: Let $f:[a,b] \to \mathbb R$ a Riemann integrable function and $M:= \sup\{|f(t)|:t \in [a,b]\}$.
If $F(x):=\int_a^x f(t) dt$, then let $x,y \in [a,b]$.
WLOG:  $x \ge y$. Then we have $F(x)-F(y)=\int_y^x f(t) dt$, hence
$|F(x)-F(y)|=|\int_y^x f(t) dt| \le \int_y^x |f(t)| dt \le \int_y^x M dt=M(x-y)=M|x-y|$.
F is Lipschitz- continuous !
Your function $S$ is  Riemann integrable !
A: Tons of other people can explain the true mathematics behind it, but I think it could be intuitive for you to try to understand where this continuity is coming from by looking at why the continuous triangle's derivative is giving you a step function. Say we have this function
$$
f(x) = |x| = \begin{cases} 
      -x & x< 0 \\
      x & 0\leq x 
   \end{cases}
$$
which is very provable to be continuous
$$
\lim_{x\rightarrow 0^-}f(x) = 0 = \lim_{x\rightarrow 0^+}f(x).
$$
Now if we differentiate the function we find that 
$$
f'(x) = \begin{cases} 
      -1 & x< 0 \\
      1 & 0> x \\ 
      &x\neq 0
   \end{cases} 
$$
which is a step function! That is why the reverse holds true: our step functions' integrals give continuous functions because we are making these piecewise linear equations that have "kinks". So long as the connection points are the same, the function will be continuous
A: You can compute the value explicitly in each interval: If $s(x) = c_k$ for $x \in [t_k,t_{k+1})$ then for $x \in [t_k, t_{k+1}]$ we
have $F(x) = F(t_k) + \int_{t_k}^x s(t) dt = F(t_k)+ (x-t_k) c_k$.
We have $F(x) = F(t_0)$ for $x \le t_0$.
A: The "new" piece of curve is not "added" as $x$ moves forward. For each $x$ we use the Riemann sums, this is why it is smoother then expected.
