If two functions share a point of discontinuity, one is not Riemann integrable resp. to the other Show that $g: [-1,1] \to \mathbb{R}$ is not Riemann integrable respect to $\alpha:[-1,1] \to \mathbb{R}$ where $g(0) = 2$ and $g(x) = 1$ everywhere else and  $\alpha(0) = -1$ and $\alpha(x) = 0$ everywhere else.
I tried calculating $U(P,g,\alpha)-L(P,g,\alpha)$ as usual, but I got $0$ because $\alpha$ is not monotonic. I'm used to work with monotonic functions there, but Apostol's book does it in general :'( This got me confused, because then it would be true that $U(P,g,\alpha)-L(P,g,\alpha)$ being zero is less than any positive $\epsilon$. Where is my reasoning incorrect? How can I show then that $g$ is not Riemann integrable resp. to $\alpha$?
Also, we are using Apostol's definition: $f$ is Riemann integrable in $[a,b]$ if there is $A \in \mathbb{R}$ such that for every $0 < \epsilon$ there is a partition $P_\epsilon$ such that for any finer partition $P$, $|S(P,f,\alpha)-A| < \epsilon$
 A: Proof 1
The integrator $\alpha$ is not monotonic but is of bounded variation. Also $\alpha$ is nondecreasing on $[0,1]$ and nonincreasing on $[-1,0]$. If $g$ is integrable with respect to $\alpha$ over $[-1,1]$ then it is necessary that it is integrable over $[0,1]$.
Let $P $ be any partition of $[0,1]$.  Hence,
$$U(P,g,\alpha) - L(P,g,\alpha) \\ \geqslant (\sup_{x \in [0,x_1]}g(x) - \inf_{x \in [0,x_1]}g(x)\,\,)(\alpha(x_1) - \alpha(0))\\ = (2-1)(0 - (-1)) \\= 1.$$
The inequality holds because every term of $U(P,g,\alpha) - L(P,g,\alpha)$ is nonnegative, and , therefore, $U(P,g,\alpha) - L(P,g,\alpha)$ must be greater than or equal to any one term.
Thus, the integrability condition $U(P,g, \alpha) - L(P,g, \alpha) < \epsilon$ cannot be satisfied for every $\epsilon > 0,$ and $g$ fails to be integrable with respect to $\alpha$ over $[0,1].$
Proof 2
Alternatively, take any partition including the points $x < 0 < y$.  Consider Riemann-Stieltjes sums $S_1(P,g, \alpha)$ with intermediate points $c_1 \in [y,0]$ and $d_1 \in [0,x]$, and $S_2(P,g, \alpha)$ with intermediate points $c_2 \in [y,0]$ and $d_2 \in [0,x]$.
If $c_1 < 0$ and $d_1 = 0$ then $S_1(P,g, \alpha) = 2(0 - (-1)) + 1 (-1-0) =1$, and if  $c_2 = 0$ and $d_2 > 0$ then $S_2(P,g, \alpha) = 1(0 - (-1)) + 2 (-1-0) = -1$.
If $g$ were integrable, then there exists some $A \in \mathbb{R}$ such that for any $\epsilon > 0$, if the partition $P$ is sufficiently fine we must have both
$$|S_1(P,g,\alpha)- A| = |1 - A| < \epsilon, \\ |S_2(P,g,\alpha)- A| = |-1 - A| < \epsilon,$$
which requires an impossible condition $A = 1$ and $A = -1$.   
