Explaining Zero Content for W.

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I was wondering if someone might be able to clarify the concept of "zero content" to me. I have the textbook definition:

Given a set Z, for any epsilon > 0, there are a finite # of rectangles such that 1) their union is a subset of Z and 2) their summed areas is less than epsilon.

But this seems very abstract to me...is there a more tangible way of explaining what zero content is?

Cheers,
W. =)
 
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If I may, what course is this for? What's the text?
 
advanced calculus, gerald b. folland is the author
 
All I can say is "welcome to advanced math". Seems abstract? Yep.

Personally almost all of this stuff seemed abstract the first time through. It's just a matter of doing exercises with them so that you start to gain a sense of why these seemingly abstract definitions have value.

So I guess my answer is "I don't know how to make it seem less abstract". Have you run into compact sets yet? The first time you see them they will probably seem about as abstract as sets of content zero, yet there is a more intuitive description of compact sets if they live in certain types of spaces (such as R^n).
 
your definition is wrong, it should read that Z is contained in the union. Think of a straight line, intuitively it has zero area and it agrees with the definition. Also the graph of a continuous function on a closed interval has zero area(zero content,zero volume, same thing) so it agrees with the definition.
 
Did you try drawing a picture?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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