Why is the region assumed to be closed in the definition of multiple Riemann integrals?

To define the multiple Riemann integrals, in (advanced or elementary) calculus textbooks, they divide the region of integration $R$ into subregions, choose an arbitrary point in each subregion and say if the following limit exists, then the function is integrable.

$$\lim_{||\Delta||\to 0}\sum_{i=1}^n f(x_i^*,y_i^*) \Delta_i A$$ where $||\Delta||$ is the norm of subdivision.

I have two questions about the definition of multiple integrals:

Question 1: At first it is assumed that $R$ is a closed bounded region. Why do they need to assume $R$ is closed (even if $f$ is bounded in $R$)? Why does this assumption make the definition easier? What problems will we face if $R$ is open?

Question 2 In one of the books, there is an additional condition: ''if no matter how we choose the grid" the limit exists. On which conditions, the way we choose subdivisions may affect the limit of the Riemann sum?

• – Kevin Jun 22 '17 at 13:29

Both of your questions have analogs for single-variable integrals. It is useful to cast your questions in the light of integrals over some interval in $\mathbb{R}$.

For Question 1

Implicit in your question is an idea that this definition makes things easier. I am not sure if that's true. What is true is that using bounded closed domains of integration is sufficient.

It is possible to define integration based on open intervals. You might even try, and see if any problems arise. One nice thing about closed intervals is that the union of $[a,b]$ and $[b,c]$ is $[a,c]$. Unfortunately, the union of the closed intervals $(a,b)$ and $(b,c)$ is not $(a,c)$. Thus using closed intervals, it is perhaps more obvious that we should expect $$\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx.$$ But in fact all three integrals are independent of the value of $f(b)$, so this is only a minor hurdle.

Perhaps it is good to note that it is very easy to define the integral of $f(x)$ over an open set $S$, using integrals over closed sets as one's fundamental definition. Choose a closed set $U$ containing $S$, and define the integral of $f(x)$ over $S$ to be the integral of $1_S (x)\cdot f(x)$ over $U$, where $1_S(x)$ is $1$ if $x \in S$ and $0$ otherwise. Going from open to closed can also be done.

For Question 2

I do not quite understand your question, but I believe you are asking when can the choice of subdivisions affect the value of the integral? For this, you need only examine any non-integrable function, such as highly discontinuous functions.

For example, see this question on Dirichlet's function $f(x)$, which is defined to be $1$ if $x$ is rational and $0$ otherwise. This function is not integrable on any interval (and has ready extension to higher variables).

• Thanks a lot. Which part of the definition implies that the value of the limit must be independent the choice of subdivisions? In other words, I want to know why other books do not mention this part of the definition. – David R Jun 26 '17 at 22:58
• That is encoded within the phrase "no matter how we choose the grid", referring to the language in your original post. It is essential to the definition of the integral, and any complete definition should include something equivalent to it. – davidlowryduda Jun 27 '17 at 1:56