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It's an excerpt from Rudin's book. I can't understand the following moments:

1) Why he considers continuous function with compact support? Why compactness is so important?

2) Why equation (3) has the meaning? Why $f$ is zero on the complement of $I^k$?

3) Why integral in (3) is independent of the choice $I^k$? It's not obvious to me.

Can anyone give an answer to my above questions? I would be very grateful.

  • $\begingroup$ I just remember one line he said in the beginning of this chapter: "the proper setting for the discussion should be the Lebesgue integral." Seeing this I simply skipped this chapter and embarked upon his Real and Complex Analysis. :) $\endgroup$ – Vim Mar 13 '16 at 16:47
  • $\begingroup$ @Vim: One still needs to learn multivariable analysis. I wouldn't choose Rudin's text for that material, but, nevertheless ... $\endgroup$ – Ted Shifrin Mar 13 '16 at 16:50

Basically, the answer to 1) is that the Riemann integral is defined only on rectangles, so you need to be able to enclose the set of points $x$ where $f(x)\ne 0$ in a giant rectangle and then integrate over that rectangle. That's what he's doing with the integral over the $k$-cell $I^k$. I'm not sure what you mean by 2); the answer to 2) should be the answer to 3). Consider any two rectangles (or $k$-cells) containing the support of $f$. On any subrectangle contained in one of those rectangles but not in the other, we have $f=0$, so integrating $f$ over that subrectangle will give $0$.

  • $\begingroup$ 1) If $\text{supp}(f)$ is compact then it's closed + bounded. Hence support of $f$ is contained in some $k$-cell. Right? $\endgroup$ – ZFR Mar 13 '16 at 16:44
  • $\begingroup$ Yes, sure. I guess I took that for granted. :) $\endgroup$ – Ted Shifrin Mar 13 '16 at 16:45
  • $\begingroup$ 2) Rudin defines equation (3). So $f$ must be zero on complement of $I^k$. If I am right how to prove it? $\endgroup$ – ZFR Mar 13 '16 at 16:46
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    $\begingroup$ That's the point of our choice of $k$-cell. All the points where $f(x)\ne 0$ are contained in the interior of the $k$-cell. Thus, outside the $k$-cell, $f$ is $0$ everywhere! $\endgroup$ – Ted Shifrin Mar 13 '16 at 16:48
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    $\begingroup$ So that you can have a compact set to work with. Saying compact support is (in Euclidean space) equivalent to saying bounded support. Although you will eventually learn measure theory and Lebesgue integration, for purposes of the Riemann integral we want to integrate over compact objects. $\endgroup$ – Ted Shifrin Mar 13 '16 at 16:59

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