Additive Property of Integrals of Step Functions

In Apostol's Calculus Volume-1 the proof of Additive Property for Integrals of Step Functions is given as an exercise that is:

$$\int_a^b[u(x)+g(x)]dx=\int_a^b u(x)dx+\int_a^b g(x)dx$$

And Integrals of step functions are defined as follows:

Let $s(x)$ be a step function defined on $[a,b]$ and let $P=\{x_0,x_1,...,x_n\}$ be a partition of $[a,b]$ such that $s(x)$ is constant on the open subintervals of $P$.

Denote by $s_k$ the constant value $$s(x)$$ takes in the kth open subinterval, so that $s(x)=s_k$ if $$x_{k-1}<x<x_k$$ for $k=1,2,...n$.

Then $$\int_a^b s(x)dx = \sum_{k=1}^n s_k(x_k-x_{k-1})$$ Where $$x_0=a$$ and $$x_n=b$$

I don't know how to prove this property using the given definition, any help will be appreciated.

• Why did you reask the same question? That is against the rules, because it may lead to people wasting their time trying to answer a question that was already answered! – Jyrki Lahtonen Jun 24 '16 at 16:45
• When I posted the question first time I didn't know how to use mathjax, so I didn't get the the expected answer but after I learned mathjax I posted the question using mathjax as separate question. – Shubhashish Jun 24 '16 at 17:07

Lemma: Suppose that $P$ is a partitition for $f$ on $[a, b]$, and $a < c < b$ (i.e, if $P = \{ x_0, x_1, \ldots, x_n\}$ with $x_0 = a, x_n = b$, $x_0 < x_1 < \ldots < x_n$, and $f$ constant on $x_i < x < x_i+1$).

Then $P \cup \{c\}$ is also a partition for $f$ on $[a, b]$.

Proof: Case 1: $c \in P$. Then $P \cup \{c\} = P$, and by hypotheses, $P$ is a partition for $f$.

Case 2: $c \notin P$. Then $c$ must be between some part of points in $P$, say $x_k$ and $x_{k+1}$. So $P' = P \cup \{c\}$ consists of $$x_0 = a < x_1 < \ldots < x_k < c < x_{k+1} < \ldots < x_n = b.$$

Because $f$ is constant on $x_k < x < x_{k+1}$, it's constant on both $x_k < x < c$ and on $c < x < x_{k+1}$. So $P'$ is a partition for $f$ on $[a, b]$.

Corollary: If $P$ is a partition for $f$ on $[a, b]$, and $Q$ is a finite set of values between $a$ and $b$, then $P \cup Q$ is a partition for $f$ on $[a, b]$.

Proof: induction on the previous lemma: add one point of $Q$ at a time.

Now consider two step functions (with finitely many steps) $f$ and $g$ on an interval $[a, b]$. Pick partitions $P$ and $Q$ for the two functions on $[a, b]$.

Now you can build a new partition $R = P \cup Q$. The corollary tells us that $R$ is partition for $f$ and is also a partition for $g$ on $[a, b]$. But it is also a partition for $f+g$ on $[a, b]$ (i.e., $f+g$ is constant on each interval of $R$). (I suggest that you draw a picture to see why this is true.)

With THIS partition, the equality you're trying to prove is actually obvious, for on any interval $r_k < x < r_{k+1}$, the functions $f$ and $g$ and $f+g$ are all constant, and the value of $f+g$ on that interval is the sum of the values of $f$ and $g$ on that interval.

• In Apostol's book the concept of continuity is defined later on so I don't understand what you mean by a set of discontinuities. – Shubhashish Jun 18 '16 at 16:46
• Let's say "the set of places where steps occur", so a function like $f(x) = 2$ for $x \le 5$ and $f(x) = 3$ for $x > 5$ has a "discontinuity" or "step" at $x = 5.$ – John Hughes Jun 18 '16 at 20:04
• I have updated my question please provide some help using the definition of Integrals of Step Functions I have given in my question. – Shubhashish Jun 20 '16 at 11:47
• See my revised answer. – John Hughes Jun 24 '16 at 15:33
• Than You, I got it now. – Shubhashish Jun 24 '16 at 16:47

Let $P=\{x_0,x_1,...,x_n\}$ be a partition of $[a,b]$ such that the step function $u(x)+g(x)$ is constant on the open subintervals of $P$. As $u(x)$ and $g(x)$ are also step functions there exists a refinement of $P$, say $\tilde{P}=\{x_{00},x_{01},...,x_{0i_1};x_{10},x_{11},...,x_{1i_1};\cdots;x_{(n-1)0},x_{(n-1)1},...,n_{(n-1)i_{n-1}};x_{n0}\}$ (where $x_{k0}:=x_k$), such that both $u(x)$ and $g(x)$ are constant in every interval of $\tilde{P}$. Now, using your notation, you have that \begin{align*} \int_a^b[u(x)+g(x)]dx &=\sum_k(u+g)_k(x_{k-1}-x_k) \\ &=\sum_{k}(u+g)_k(x_{(k-1)0}-x_{(k-1)1}+x_{(k-1)1}-\cdots-x_{(k-1)i_{k-1}}+x_{(k-1)i_{k-1}}-x_{k0})\\ &=\sum_{k,j}(u+g)_k(x_{(k-1)(j-1)}-x_{(k-1)j})\\ &=\sum_{k,j}[u_{kj}+g_{kj}](x_{(k-1)(j-1)}-x_{(k-1)j})\\ &=\sum_{k,j}u_{kj}(x_{(k-1)(j-1)}-x_{(k-1)j})+\sum_{k,j}g_{kj}(x_{(k-1)(j-1)}-x_{(k-1)j})\\ &=\int_a^bu(x)dx+\int_a^bg(x)dx \end{align*} The underlying idea for the demostration is easy, but when formalizing it turns out to be a little tedious to write.

• In the refinement set isn't $x_{n0}$ the last point. – Shubhashish Jun 20 '16 at 13:45
• @user348631 Absolutely. Sorry for the little flaw, editing right now – user335721 Jun 20 '16 at 13:48
• what does $$\sum_k$$ and $$\sum_{k,j}$$ mean? – Shubhashish Jun 20 '16 at 13:57
• @user348631 It means "sum everywhere where it makes sense to". Usually one writes $\sum_i$ as a lazy version of $\sum_{i=n}^m$ when the subindex and superindex are assumed to be understood without mention, just like in the Einstein summation convention. In this case I did so cause it would be tedious to write all the indexes (notice that the $j$ index runs till $i_{k-1}$, which depends of $k$.) – user335721 Jun 20 '16 at 14:24

I'm assuming you want the formula for step functions. The general formula follows from this, using the definition of the integral.

So, let $u=\sum _k u_i\chi _{[x_i-x_{i-1}]}$ and $v=\sum _k v_i\chi _{[x_i-x_{i-1}]}$. Then

$$\int_{a}^{b}(u+v)(x)dx=\int_{a}^{b}(u(x)+v(x))dx=\int_{a}^{b}\left ( \sum _i u_i\chi _{[x_i-x_{i-1}]}(x)+\sum _i v_i\chi _{[x_i-x_{i-1}]}(x) \right )dx=\int_{a}^{b}\sum _i \left ( u_i\chi _{[x_i-x_{i-1}]}(x)+ v_i\chi _{[x_i-x_{i-1}]}(x)\right )dx=\int_{a}^{b}\sum _i (u_i+v_i)\chi _{[x_i-x_{i-1}]}(x)dx=\sum _i (u_i+v_i)\Delta x_i=\sum _i u_i\Delta x_i+\sum _i v_i\Delta x_i=\int_{a}^{b} u(x)dx+\int_{a}^{b} v(x)dx$$

The first equality is definition of the sum of two functions, the second is substitution, the third is algebra, the fourth definition of the integral, the fifth is algebra and the sixth again the definition of the integral.

• Have you used my definition for the Integral of step function, I don't understand your terminology. – Shubhashish Jun 20 '16 at 13:18
• Yes I wrote the formulas for the step functions in a slightly different but more compact way. – Matematleta Jun 20 '16 at 14:00
• What does $$u=\sum_k u_i\chi _{[x_i-x_{i-1}]}$$ mean in simple notation? – Shubhashish Jun 20 '16 at 14:05