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?
Best Answer
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 !