# Some Analysis proofs (complete, just need a check)

1. Nov 7, 2008

### TaylorWatts

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.

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.

2. Nov 7, 2008

### 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.