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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.


marked as duplicate by C. Falcon, choco_addicted, Alex M., Lee Mosher, user223391 Jun 24 '16 at 22:19

This question was marked as an exact duplicate of an existing question.


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.

  • $\begingroup$ In the refinement set isn't $x_{n0}$ the last point. $\endgroup$ – Shubhashish Jun 20 '16 at 13:45
  • $\begingroup$ @user348631 Absolutely. Sorry for the little flaw, editing right now $\endgroup$ – user335721 Jun 20 '16 at 13:48
  • $\begingroup$ what does $$\sum_k$$ and $$\sum_{k,j}$$ mean? $\endgroup$ – Shubhashish Jun 20 '16 at 13:57
  • $\begingroup$ @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 $k$.) $\endgroup$ – user335721 Jun 20 '16 at 14:24

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