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Real Analysis I: Hints for Problems of Chapter 3

Section 3.3

1. 

  if x is rational
if x is irrational

Let be any partition of [0, 1]. Then in any subinterval , we can pick a
rational number and an irrational number so that and . Thus for any
partition P, we have , while



. f is not Riemann integrable since   while .

2. Let
Let ε > 0 be given. Let be any partition of [0, 1] such that the length of the largest
sub interval is less than ε. (It is a matter of choosing N so large that ε > 1/N.) Suppose 1/2 is in the jth
subinterval . On this subinterval while . On all other subintervals we have and
(why?). We now have


and

Subtracting the two equations we get



Therefore, f is Riemann integrable by Lemma 3 (Page 89).

3. Suppose f is Riemann Integrable and f(x) ≥ 0 for all x ∈ [a, b]. (a) To show , let
. Then for each subinterval of this partition, we have
. (Explain why should be ≥ 0.) But then and hence . By
definition and hence .

(b) Suppose f(x) ≥ 0, f is continuous and . We need to show f(x) = 0 for all x. Suppose
not. Then there exists a number c in [a, b] such that f(c) > 0. Since f is continuous, there exists a delta > 0
such that f(x) > 0 for all x in [a, b] and |x − c| ≤. We can choose smaller, if necessary, so that c − and
c + are in [a, b]. Let . Then is a partition of
[a, b].( Draw a number line and show this partition.) Since f(x) ≥ 0 on the first and the third subintervals
we have, and . On the second subinterval , the function is continuous. Hence it
attains its maximum and minimum. On this interval (by the choice of) f(x) > 0 for all x. In particular
. Thus



Since is greater than or equal to any lower sum we see that . This contradicts the
assumption.
(Note: Once we assume that f(c) > 0 for some c, and f(x) ≥ 0 for all x, then is the area of a region
under the graph and hence it must be positive. This is what we proved above, a seemingly trivial statement!!)

4. Let f(x) = 3x on [0, 1]. and let ε > be given.

(a) Let = ε/3. Then for all x and for c in [0, 1], |x − c| ≤ implies |f(x) − f(c)|≤ . Now choose a
positive integer N large enough so that 1/N < and let



For each subinterval, we have and . (Draw the graph of f (x) = 3x and use N = 8
to see this or simply observe that the function is increasing.) But then



and



Hence as required.

(b) Let k be any integer and let . Let . Then is in the subinterval
. Note then that and . We form the Riemann sum


.
Thus by Corollary 3.3.2 (page 91), we have



5. Since f is continuous on [a, b], by Corollary , where
is a Riemann sum. Any Riemann sum can be divided into parts where f is positive and f is negative. On
the positive parts is the area of the rectangle whose height is and width is.
On the negative parts is negative one times the area of the rectangle whose height is
and width is . Thus the Riemann sums are the sum of the areas of the rectangles above the x-axis
minus the sum of the areas of the rectangles below. Passing to the limit, we conclude that can be
interpreted as the sum of the areas above the x-axis minus the areas below. The Theorems mentioned make
sense because areas under graph of functions satisfy these properties . (You may want to draw graphs for each
of the theorems and the corollary.)

7. Note that f(x) = x2 is increasing on [1, 2]. Thus for any partition of [1, 2],
and . In other words, the lower sums are obtained by using the left endpoint of the
subintervals while upper sums are obtained by using the right endpoints.

8. First observe that for any x and c in [1, 3], we have x ≥1 and c ≥1. Hence



. Next note that the function is decreasing and therefore for any partition of [1, 3],
and . In other words, the lower sums are obtained by using the left endpoint of the
subintervals while upper sums are obtained by using the right endpoints. Choose N = 102 and let the above
partition be chosen so that and so on. (Note then that
= 1+2n/N = 3 as required!) Compute and and subtract .

9. Show that for all x ∈ [0, 1] and integrate.

13 First note if x < c, then . If f is continuous on [a, b] then it is bounded:
say |f(t)| < M for all t in [a, b]. Let ε > 0 be given and let= ε/M. The for all x and all c in [a, b], is
and x < c, then



(Explain each equality and inequality in the above argument.) Give the argument for the case if x > c.

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