N-th dimensional Riemann integral

[nightingale123]

Hello I have a question regarding something we wrote in class today.
Let $A$ be a bounded subset of $R^n$, let $f,g:A\to \mathbb{R}$ be integrable functions on A.
$a)$ if $A$ has a volume and $\forall x \in A :m\leq f(x) \leq M$ then $mV(A)\leq \int_{A}f(x)\leq MV(A)$

this kinda surprised because I've always that that we could only integrate on sets who have a volume (circles squares etc.).

Could somebody give me an example of a set $A$ which does not have volume and a function $f:A\to \mathbb{R}$ which is integradable

 

[mathman]

If A has no volume, presumably volume=0, then any integral of a function A-> R would be 0.

 

[mathwonk]

I don't know what precise definitions you are using but I see it this way. The Riemann integral of a function can be defined without regard to a specific set, by approximating the function by step functions. In this way one always regards every function as defined on the whole eucliden space, usually with value zero at most points. For exampe the function with value 1 on the closed unit interval [0,1] can be regarded as having domin the whole real line by setting it equal to zero outside that interval. Thus even though it has domin the whole real line, which does not have a finite length, it still has a finite integral equal to 1.

If we choose any other subset A of the real line, we cn define the integral of f over A, by first multiplying f by the "characteristic function" of A, and then integrating the product. This results in leaving the values of f the same at points inside A, but chopping them off to zero at points outside A. Now as long as A is a set containing the unit interval, this will not change f at all nor its integral, and thus f will still be integrable over any set, no matter how complicated, as lon g as the set contains the unit interval

So the point is the integrability of f depends more on the set where f has non zero values than it does on the nature of the set A you say you are integrating over. So take a function f in the plane which equals 1 everywhere inside or on the unit circle and equals zero outside the unit circle. Now cook up some larger set A that contains the unit circle but that has such a ragged border so that its area is not defined. We can still integrate f over that awful set A because f has value zero at the worst parts of A, and the integral still has the same value, namely pi. But since A has no well defined area, we cannot compare the integral of f to the area of A.

does this make sense?

 

[Svein]

A really hairy example exists, see https://en.wikipedia.org/wiki/Swiss_cheese_(mathematics).

A short explanation of how to construct one:

1. Start with a disc with radius 1
2. List all points in the circle where both coordinates a rational (this is a countable set)
3. Start with point 1. Punch a hole in the disc. The hole should be wholly contained in the disc and the radius should be ≤½
4. Continue punching holes ad infinitum. Each hole should be wholly contained in the remains of the disc and the radius of hole n should be ≤1/2ⁿ

The remains of the disc then contains no open regions. The area of hole n is ≤π⋅1/2²ⁿ, so the total area of the holes will be ≤π⋅∑1/2²ⁿ=π/3.
Thus the remains of the disc (with no "interior points") has an area of at least π⋅2/3.

 

[mathwonk]

I believe Svein is giving an example of a set with no area. In the sense of Riemann integration, a set has "content" iff its inner and outer contents are equal, i.e. its approximations by means of inner and outer unions of rectangles. Since his example has no interior, it contains no rectangkles at all, so the inner content is zero. But apparently the outer content must be positive, so the content oif this set is undefined. Nonetheless, if we take a function that equals zero on this whole set, then it has an integral over this set equal to zero.

I am an outsider in this subject and have not worked too hard on this, so might be off somewhere, but i did think about it a bit, and reviewed Loomis and Sternberg on the theory of integration via content, which I recommend to you.

 

[mathwonk]

Nightingale, I hope you are seeing this. If so, would you please comment? Even if it is all "Greek" we may be able to clarify.

 

[nightingale123]

I apologies for not replying sooner I was kinda depressed because I did bad on my midterm.

However I have though about this questions since then. We defined that a set has a volume when the volume of the boundary of the set is 0 in other words. $V(\partial A)=0\Leftrightarrow \exists V(A)$ and we also showed that if we define $A:=\mathbb{Q}\cap[0,1]^2$ then $A$ does not have volume as $V(\partial A)=V([0,1]^2)=1\neq 0$. Then I though of this example but I'm not sure if it works. I though of choosing $\forall x : f(x)=0$. And since $A,f$ are bounded and f is continuous that makes f integrable. However this is a really dumb example and I'm not sure if it even works

 

[mathwonk]

that is a perfect example! well done. (You might want to write Q^2 instead of Q, since your example seems to be in 2 dimensions.)