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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: enter image description here

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

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?

rain
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  • If you consider the two variable case where S(X) = c0 for x<0 and c1 for x>=0. If you do an indefinite integral you get S(X) = c0x+d0 for x<0 and c1x+d1 for x>=0 (d0 and d1 are constants). I don't see how this is continuous for all d0 and d1. Sure there will always exist d0 and d1 for which the function will be discontinuous. I think the indefinite integral CAN be continuous for some values of the constants of integration (d0 and d1) – Shreyas Pimpalgaonkar Jun 29 '18 at 04:02
  • @ShreyasPimpalgaonkar Hi, I will update a picture from Apostol that says that it must necessarily be continuous. I'm not sure why though. – rain Jun 29 '18 at 04:03
  • @ShreyasPimpalgaonkar I added they update, let me know what you think? thanks. – rain Jun 29 '18 at 04:12
  • @dxiv I just read through the page you sent about Darboux's theorem (I've heard of it from analysis) but I'm not sure how that explains how the integral of a step function is continuous / cannot have jump discontinuities? Could you please elaborate, much appreciated. Thank you. – rain Jun 29 '18 at 04:24
  • Okay I got it. So you have to see that a is the same for every piece here. Put a = -10000 (basically smaller than t0 and do your calculations) You will then see. Then put a = anything and then do the same. – Shreyas Pimpalgaonkar Jun 29 '18 at 04:24
  • A more clear answer https://math.stackexchange.com/questions/1730049/is-the-indefinite-integral-of-a-piecewise-continuous-function-a-continuous-funct – Shreyas Pimpalgaonkar Jun 29 '18 at 04:26
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    @rain Sorry, misread it. Comment removed, see Is an integral always continuous? instead. – dxiv Jun 29 '18 at 04:28
  • @ShreyasPimpalgaonkar Hi so when I integrate every piece I use a (lower bound) for all of them, and integrate up to x everytime? – rain Jun 29 '18 at 04:29
  • @dxiv Hello thanks for the reply. I think what would help me is considering how they computed the integral of the piecewise function T(X) above. I'm not sure how they got x 0<=x<2 and 4-x for 2<=x<=4. I'm not sure how to/what it means to compute the indefinite integral of a piecewise function. So if you could just briefly explain how they came up with that, would really help me for understanding whats happening – rain Jun 29 '18 at 04:32
  • @dxiv The link is very helpful. We know that a step function such as the one above is Riemann integrable because it is continuous almost everywhere (its set of discontinuities form a set of measure zero). Is that correct? – rain Jun 29 '18 at 04:42
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    @rain Intuitively, it's obvious that the area under the step function varies continuously with no jumps. It's just that at each jump of the step function the rate of increase (or decrease) of that area changes. To compute $T(x)$ note that it's linear $T(x)=c_k x + C_k$ on each interval $[t_k, t_{k+1}]$, then determine the constants $C_k$ such that $T(x)$ is continuous, which you know it must. – dxiv Jun 29 '18 at 04:42
  • @dxiv I see I just wasn't sure if the computation was done without using the assumption that it was continuous. That is I thought that maybe the computation would help me show that it is continuous. Thanks – rain Jun 29 '18 at 04:45

4 Answers4

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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 !

Fred
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  • I just proved this thanks for the help. Is a step function Riemann integrable because it is continuous almost everywhere (its set of continuities form a set of measure zero)? Thanks. – rain Jun 29 '18 at 07:29
  • Yes, if $f:[a,b] \to \mathbb R$ is bounded , then $f$ is Riemann integrable $ \iff$ $f$ is continuous a.e. – Fred Jun 29 '18 at 07:58
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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

wjmccann
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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 $x \le t_0$.

copper.hat
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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.

user123124
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  • How does a Riemann sum (it is just a sum dealing with values of function at finite number of points) make it smooth? You need to explain clearly what you are saying. – Paramanand Singh Jan 21 '21 at 14:28