# Definition 10.3 from PMA Rudin

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.

• 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. :) – Vim Mar 13 '16 at 16:47
• @Vim: One still needs to learn multivariable analysis. I wouldn't choose Rudin's text for that material, but, nevertheless ... – 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 $0$.
• 1) If $\text{supp}(f)$ is compact then it's closed + bounded. Hence support of $f$ is contained in some $k$-cell. Right? – ZFR Mar 13 '16 at 16:44
• 2) Rudin defines equation (3). So $f$ must be zero on complement of $I^k$. If I am right how to prove it? – ZFR Mar 13 '16 at 16:46
• 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! – Ted Shifrin Mar 13 '16 at 16:48