1 Which sets have volume 0?

Transcription

1 Math 540 Spring 0 Notes #0 More on integration Which sets have volume 0? The theorem at the end of the last section makes this an important question. (Measure theory would supersede it, however.) Theorem set R n has volume zero if and only if for every " > 0 there is a nite covering of by rectangles with total volume less than ": This is worked example 8. on page 487, so I will omit the proof in these notes. The only if part was a lemma in notes #9. (Note the di erence between this and the de nition of a set of measure 0: In the latter case, you can allow an in nite covering, so long as it is countable.) We can now list some sets with zero volume, and ask some interesting questions about volume zero and measure zero.. nite set of points. This should be obvious; if there are m points, and " > 0; nd a box of volume less than " containing each point. m. The image of a smooth curve : [a; b]! R n : (This is implied by a homework problem, which is more di cult than most.) 3. What about the image of a continuous curve : [a; b]! R n? 4. What about the image of f (x) = x; sin x over (0; ]? 5. (see page 454, # 5 and 6): Must the boundary of a set have measure zero? Must the boundary of a set of measure zero have measure zero? Suppose that has no volume. Can its have volume?

2 Properties of the integral These are listed in Theorem They all follow from a similar property for sums. The one needing discussion is: Theorem Mean Value Theorem for Integrals: Suppose that f :! R, with R n bounded, f is continuous, and the boundary of has volume zero. Suppose that is compact and connected. Then R f exists, and there is an x 0 such that f = f (x 0 ) V () : Proof: The last theorem of the previous notes shows that R f exists. To get the mean value theorem, let m = inf ff (x) j x g and M = sup ff (x) j x g : Then by the de nitions of upper and lower sums, we see that mv () < R f < R f MV () ; or m < < M: Because is compact and f is continuous, m and V () M are values taken on by f in : Hence by the intermediate value theorem for R a continuous function on a connected set, there is an x 0 such that f (x 0 ) = f ; V () which implies the theorem. 3 Improper integrals We have only de ned R f in the case where is bounded and f is bounded. If either of these conditions is not satis ed, then R f has not been de ned. We rst deal with the case where is unbounded. We also assume that f is nonnegative. De nition 3 Suppose that R n and f :! R and f is bounded. Suppose also that f (x) 0 for all x : Suppose that for every n; B n = [ n; n] [ n; n] [ n; n] : (This will be called the n - cube.) Suppose that for each n, R \B n f exists. Suppose nally that lim f n! \B n exists. Then we de ne R f to be this limit. Next we consider the case where f may be unbounded. We still assume that f 0: In this case, we let f (x) if x and f (x) M f M (x) = 0 if f (x) > M:

3 De nition 4 If f M is integrable over for each M; and lim M! R f M exists, then we de ne R f to be this limit. When we say integrable over in this de nition, we mean according to the previous de nition, which allows to be unbounded. Finally, we remove the requirement that f 0: Let f (x) if x and f (x) 0 f + (x) = 0 if x and f (x) < 0 and f (x) = f (x) if x and f (x) 0 0 if x and f (x) > 0 Then for any x ; f (x) = f + (x) f (x) : We de ne R f to be R f + R f ; assuming both of these exist according to the de nitions above. 3. The last de nition is important! This de nition leads to an interesting conclusion. Let = ( ; ) R; and on ; let f (x) = x: Does R f exist? You might think so, because when n = ; B n = [ n; n] ; and n fj [ n;n] = 0 n for each n: But and so R f + does not exist. 3. n example in R : n f + = lim f + = n! n n 0 f + = n ; Therefore, R f does not exist. s far as I have seen, the text gives no examples, at least in this chapter, of improper integrals which exist when n > : Here is an example for n = : Let = f(x; y) j x + y g : For (x; y) ; let f (x; y) = ( if (x; y) 6= (0; 0) (x +y ) 3=4 0 if (x; y) = (0; 0) : 3

4 To show that this integral exists I will jump ahead in the book and use the theory of change of variables in sections 9.3 and 9.4, which shows that this integral exists if and only if 0 g exists, where 0 is the rectangle in R given by and in 0 ; 0 = f(r; ) j 0 r ; 0 g ; g (r; ) = pr if r 6= 0 0 if r = 0: (Recall from calc III that you do an integral in polar coordinates by setting R R R R R R f (x; y) dxdy = f (r cos ; r sin ) rdrd: Then dxdy = R R r dr d = R R pr dr d: (x +y ) 3=4 0 r 3= 0 This is all justi ed in Chapter 9. ) Note that g does not depend on ; and that R g is an improper integral. You 0 evaluate this integral in the usual way: pr if r M g M (r; ) = 0 if r < M In chapter 9 the usual theory of iterated integral is developed, as a practical way of evaluating integrals in R n : In this case it turns out that! r! g M = pr dr d = 4 4 : 0 0 M M Then we see that lim M! g R 0 M = 4; which is therefore the value of R g and R f 0. This example is interesting, because if R is the x - axis, then fj (x; y) = jxj 3= : But R jxj 3= := lim "!0 + R " + R jxj 3= " [x] 3=, which does not exist. So it s surprising that f is integrable. In some sense the reason is that the area of a small square, say of side is ; which is (a lot) less than the length of its sides if is very small. So in a partition, the areas of the squares will be less than the lengths of the partitioning line segments on the x axis. This is similar to the example of a building which can be lled on the inside with paint but is too big to paint on the outside. I ll discuss this in class! 4

5 3.3 Examples of improper integrals. There are several examples on page 465. These are all in R : Here are two of them.. R log x dx: Use integration by parts: 0 We use l Hôpital s rule: " log x dx = x log xj " " x dx = " log " ( ") : x So the nal answer is : log " lim " log " = lim "!0 + "!0 =" = lim =" "!0 = lim " = 0: "!0 ". R dx : log x which diverges. M M log x dx > x dx 3.4 Other types of convergence. Since improper integrals involve limits, if the limit exists then the integral is called convergent. De nition 5 n integral R f is absolutely convergent if R jfj exists. De nition 6 n integral R f is called conditionally convergent if it is convergent but not absolutely convergent. Example 7 Let f (x) = sin x for x < : The text shows that R f is conditionally x convergent. 4 Homework, due March 6.. Page 454, #. problem further down is related to this.. Page 454, #. 3. page 457, # 5 5

6 4. page 490. # pg. 49, #. (The "Gamma function" is important.) 6. pg. 494, # 38. Explain why the function in this problem is integrable, while the characteristic function of the rationals in [0; ] is not integrable. Hint: theorem we have studied is relevant. 7. Prove that if f : [0; ]! R is continuous, then the graph of f is a set of zero volume in R : (Recall that the graph of f is f(x; y) j 0 x and y = f (x)g : Hint: I found it helpful to think of this graphically. 8. Prove that if = ( ; ) is a smooth curve in R (so that : [a; b]! R ), then V ( ([a; b])) = 0: Hint: Boundedness of 0 is important. nd it would be enough just to have continuous. nd having D 6= 0 is not important. nd the solution to #7 is a helpful guide. 6