TaylorWatts
Nov7-08, 09:44 PM
1. The problem statement, all variables and given/known data
Suppose f >= 0, f is continuous on [a,b], and {integral from a to b} f(x)dx = 0.
Prove that f(x) = 0 for all x in [a,b]
2. Relevant equations
3. The attempt at a solution
Suppose there exists p in [a,b] s.t. f(p) > 0.
Let epsilon = f(p) / 2 > 0.
The continuity of f implies:
there exists & > 0 s.t. |x-p| < & implies |f(x) - f (p)| < f(p) / 2.
Then there exists [x1, x2] contained inside both [a,b] and [p-&, p+&] AND p is an element of [x1, x2].
Then if x is in [x1, x2], |f(x) - f(p)| < f(p) / 2.
Which means for all x, f(x) > f(p) / 2.
And since [x1, x2] is compact and f is continuous, there exists q in [x1, x2] s.t. f(q) = inf of f(x). But since f(q) >= 0, that means:
Sup Sum i=1 to n of mi * delta xi > 0 (mi are the respective infs).
Therefore integral f(x)dx from a to b > 0.
Which is a contradiction.
1. The problem statement, all variables and given/known data
Let f(x) = 0 for irrational x, let f(x) = 1 for rational x, prove that f is not Riemann integrable on [a,b] for any a < b.
2. Relevant equations
3. The attempt at a solution
Let [x1, x2] be in the set of partitions of [a,b].
Then there exist c in [x1, x2] s.t. f(c) = 1 (between any two real numbers is a rational number, since the rational numbers are dense in the reals).
Also, there exists d in [x1, x2] s.t. f(d) = 0 (same logic for irrationals).
=> mi = 0 and Mi = 1. (mi = inf, Mi = sup of the respective interval).
therefore, inf sum from i=1 to n Mi * delta xi > sup sum from i=1 to n mi * delta xi.
therefore, f is not Riemann integrable.
1. The problem statement, all variables and given/known data
Suppose f is a bounded real function on [a,b] and f^2 is Riemann integrable. Does it follow that f is Riemann integrable? Does the answer change if we assume f^3 is Riemann integrable?
2. Relevant equations
3. The attempt at a solution
No. Let f(x) = 1 if rational, -1 if irrational.
Then f^2(x) = 1 for all x.
ie (mi)^2 = (Mi)^2 then just follow the same definition as in the previous problem. But f is not integrable because mi < Mi (always).
It changes if f^3 is Riemann integrable (it's now true). Because (mi) ^ 3 = (Mi) ^ 3 if and only if mi = Mi.
Suppose f >= 0, f is continuous on [a,b], and {integral from a to b} f(x)dx = 0.
Prove that f(x) = 0 for all x in [a,b]
2. Relevant equations
3. The attempt at a solution
Suppose there exists p in [a,b] s.t. f(p) > 0.
Let epsilon = f(p) / 2 > 0.
The continuity of f implies:
there exists & > 0 s.t. |x-p| < & implies |f(x) - f (p)| < f(p) / 2.
Then there exists [x1, x2] contained inside both [a,b] and [p-&, p+&] AND p is an element of [x1, x2].
Then if x is in [x1, x2], |f(x) - f(p)| < f(p) / 2.
Which means for all x, f(x) > f(p) / 2.
And since [x1, x2] is compact and f is continuous, there exists q in [x1, x2] s.t. f(q) = inf of f(x). But since f(q) >= 0, that means:
Sup Sum i=1 to n of mi * delta xi > 0 (mi are the respective infs).
Therefore integral f(x)dx from a to b > 0.
Which is a contradiction.
1. The problem statement, all variables and given/known data
Let f(x) = 0 for irrational x, let f(x) = 1 for rational x, prove that f is not Riemann integrable on [a,b] for any a < b.
2. Relevant equations
3. The attempt at a solution
Let [x1, x2] be in the set of partitions of [a,b].
Then there exist c in [x1, x2] s.t. f(c) = 1 (between any two real numbers is a rational number, since the rational numbers are dense in the reals).
Also, there exists d in [x1, x2] s.t. f(d) = 0 (same logic for irrationals).
=> mi = 0 and Mi = 1. (mi = inf, Mi = sup of the respective interval).
therefore, inf sum from i=1 to n Mi * delta xi > sup sum from i=1 to n mi * delta xi.
therefore, f is not Riemann integrable.
1. The problem statement, all variables and given/known data
Suppose f is a bounded real function on [a,b] and f^2 is Riemann integrable. Does it follow that f is Riemann integrable? Does the answer change if we assume f^3 is Riemann integrable?
2. Relevant equations
3. The attempt at a solution
No. Let f(x) = 1 if rational, -1 if irrational.
Then f^2(x) = 1 for all x.
ie (mi)^2 = (Mi)^2 then just follow the same definition as in the previous problem. But f is not integrable because mi < Mi (always).
It changes if f^3 is Riemann integrable (it's now true). Because (mi) ^ 3 = (Mi) ^ 3 if and only if mi = Mi.