e>U(P1_e\union P2_e,f,g2)-L(P1_e\union P2_e,f,g2)
which proves that \int_a^b f d(g1+g2) exists.
We also get the equality from U(P,f,g1+g2)=U(P,f,g1)+U(P,f,g2).
2. This follows from #3.
3. mesh-integrable=> integrable: Given e>0, we must show that there is partition P_e such that e>U(P_e,f)-L(P_e,g).
Since f is mesh-integrable, choose a partition P, whose mesh is less than d>0, with e/2>|R(P,f,P*)-I|, for any choice of P*. So by taking sup and inf over all such P*, we have e/2>=|U(P,f)-I|, and e/2>=|U(P,f)-I|, thus e>U(P,f)-L(P,f), so P works. Since e/2>=|U(P,f)-I|, and e>U(P,f)-\int_a^b f, we have 3e/2>|I-\int_a^b f|, for any e>0, thus I=\int_a^b f.
next integrable=>mesh-integrable: Let |f| on [a,b] be bounded by M.
Given e>0, we need to find a d>0 such that all P with mesh less than d have e>|R(P,f,P*)-\int_a^b f|. We know from integrability that that there is a P_e such that e>|R(P_e,f,P*)-\int_a^b f| for all choices of P*. We know this also occurs for any P refining P_e, because we have e>U(f,P)-L(f,P). We need it for all P with mesh less than d.
Let's find how much U(P \union P_e,f) could differ from U(P,f). Let's try to add a single point t to P, say t is between x_{i-1} and x_i. Then in the defn of U(P\union {t},f) all of the terms are the same as those terms in U(P,f), except for a single term
(1)=(sup(f(x):x \in [x_{i-1},x_i])*(x_i-x_{i-1})
which is now replaced by two terms
(2)=(sup(f(x):x \in [x_{i-1},t])*(t-x_{i-1}) +(sup(f(x):x \in [t,x_i])*(x_i-t).
we see that |(1)-(2)| is at most 2*M*(x_i-x_{i-1} which is in turn at most 2*M*mesh(P). Thus
2*|P_e|*M|*mesh(P)>=U(P \union P_e,f)-U(P,f)|.
We can apply the same reasoning to the lower sums, and conclude that
e+4*|P_e|*M*mesh(P)>=U(P,f)-L(P,f).
So if d=e/4M|P_e|>=mesh(P) we get 2e>=U(P,f)-L(P,f) which is good enough.
Rudin #5: No take f(x)=1 for x rational and -1 for x irrational. Then f^2 is Riemann integrble but f is not. yes for f63, because the real-valued function x^(1/3) is continuous, use Theorem 6.11.
Rudin #8: Since f is positive we must show that \int_1^N f is bounded independent of N exactl y when \sum_{n=1}^N f(n) is bounded independent of N.
Consider the partition P of [1,N] consisting of the integers from 1 to N. Because f decreases monotonically we have
L(P,f)=\sum_{n=2}^N f(n), U(P,f)=\sum_{n=1}^{N-1} f(n).
Thus \sum_{n=1}^{N-1} f(n)>\int_1^N f>\sum_{n=2}^N f(n).
If \sum_{n=1}^\infty f(n)=A coverges, then A>=\int_1^N f.
If \int_1^\infty f=B converges, then B>=\sum_{n=2}^N f(n).