1. (2 point) Show that if f is integrable on [a,b], then for any sub-interval [c,d]⊂[a,b], f is integrable on [c,d].
2. (2 point) If f is a continuous non-negative function on [a,b], and ∫abfdx=0, then f(x)=0 for all x∈[a,b].
3. (3 point) Let f:[0,1]→R be given by
f(x)={0sin(1/x)if x=0if x∈(0,1].
And let α:[0,1]→R be given by
α(x)={0∑n∈N,1/n<x2−nif x=0if x∈(0,1].
Prove that f is integrable with respect to α on [0,1].
Hint: prove that α(x) is continuous at x=0.
4. (3 point) Let p,q>0 be positive real numbers, such that 1/p+1/q=1. Prove that,
if f,g are bounded real functions on [a,b] that are Riemann integrable, then
∫fgdx≤[∫∣f∣pdx]1/p[∫∣g∣qdx]1/q
Hint:
(a) If u≥0,v≥0, then uv≤pup+qvq
If you cannot prove this, you may assume it and proceed (no points taken off). If you want to prove it, you may fix u and let v vary from 0 to ∞, and watch how pup+qvq−uv change, and obtain that at the minimum the quantity is still non-negative.
(b) If f,g are non-negative Riemann integrable functions on [a,b], and
∫fpdx=1,∫gq(x)dx=1
Show that ∫fgdx≤1.
Suggested reading:
1. Ross theorem 32.7, if a function f is Riemann integrable on [a,b], then as 'mesh-size' of a partition goes to 0, the gap U(P,f)−L(P,f) tends to 0.
2. There is a 'Lebesgue criterion for Riemann integrability', see here. A weaker version that avoids introducing Lebesgue measure is the following: if f:[a,b]→R is bounded and real, and f has countably many discontinuities, then f is Riemann integrable. You can try to prove this using a similar strategy to Theorem 6.10 in Rudin.