Theorem 6.19 of Baby Rudin

Solution 1:

As $f\in\mathscr{R}(\alpha)$, by Theorem 6.6 there exists a partition $P$ consisting of $x_i\in[a,b]$ such that

$$U(P,f,\alpha)-L(P,f,\alpha) < \epsilon\ . $$

But $Q$ defined by the values $\varphi(y_i)=x_i$ is a partition of $[A,B]$ and by the equality (38) we have

$$U(Q,g,\beta)-L(Q,g,\beta) < \epsilon\ .$$

Thus by the other direction of Theorem 6.6 we have that $g\in\mathscr{R}(\beta)$.

To see that they are equal, you just need to see that

$$\sup_{P}L(P,f,\alpha) = \sup_Q L(Q,g,\beta)\ .$$

Let $P=\{x_i\}$ be a partition of $[a,b]$ and define $Q=\{\varphi^{-1}(x_i)\}$. This is a partition of $[A,B]$ so we have

$$ L(P,f,\alpha) = L(Q,g,\beta)\leq \sup_{Q}L(Q,g,\beta)\ .$$

Now we take the $\sup$ over all such $P$ to see

$$\sup_{P}L(P,f,\alpha) \leq \sup_Q L(Q,g,\beta)\ .$$

Conversely, suppose $Q=\{y_i\}$ is a partition of $[A,B]$ and we can define $P=\{\varphi(y_i)\}$ which is a partition of $[a,b]$. Then we have

$$L(Q,g,\beta)=L(P,f,\alpha)\leq \sup_{P} L(P,f,\alpha)\ .$$ Taking $\sup$ over all such $Q$ gives

$$\sup_{Q}L(Q,g,\beta) \leq \sup_P L(P,f,\alpha)\ .$$