Some Analysis proofs (complete, just need a check)

[TaylorWatts]

Homework Statement

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]

Homework Equations

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.

Homework Statement

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.

Homework Equations

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.

Homework Statement

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?

Homework Equations

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.

 

[Dick]

Without checking every detail of every line, yes, you have the right idea on all of these. 1) if f(x) is positive anywhere then the integral must be positive because x is positive on some interval. 2) The max values are 1 and the min values are 0 on any interval. Upper sums and lower sums can't have a mutual limit. 3) For f^2. Not true. You have exactly the right counter example. For f^3, right, the cube function has a continuous inverse.