This is a follow-up to this question.
Here I ask to check my work and improve the final part that I feel is missing some important steps:
So to prove what is asked for in the link above (I am not copy-pasting the question because I fear it might become too lengthy but I can do it if it is hard to follow my answer) I proceed as follows:
To see why $\int^a_bf$ exists according to the first definition I choose as a partition the set $ P_\epsilon = \{ a,c,b \}$, we notice that $\sum(f,P_\epsilon, \alpha) = \sum_{k = 1}^n f(t_k) \Delta\alpha_k = 0$ because $\alpha(b) - \alpha(c) = 0$ and $f(t_1) = 0$ no matter which $t_1$ is chosen. Also any refinement $P_r$ of $P_\epsilon$ still has $\sum(f,P_r, \alpha) = 0$, we can see this by noticing that to get a refinement one needs to add a point $y$ between $a$ and $c$ or between $c$ and $b$. But $\forall y$ chosen $\alpha(y) - \alpha(a) = \alpha(b) - \alpha(y) = 0$.
So choosing $\int^a_bf$ = 0 satisfies the first definition.
Now to prove that $\int^a_bf$ does not exist according to definition 2 one must prove that there is some $\epsilon >0$ s.t. $\forall \delta >0$ there is a partition $P$ with $||P||< \delta$ s.t.
$$\left|\int_a^b f-\sum(f,P, \alpha)\right|\ge \epsilon$$.
We can always obtain a partition $P$ such that $\left|\int_a^b f-\sum(f,P, \alpha)\right| \neq 0$ noticing that we can make $\sum(f,P, \alpha)$ = 1 or $\sum(f,P, \alpha) = 0$ by omitting or including the point $c$ from the chosen partition and adding additional points to satisfy $||P|| < \delta$.
So no $\int^a_bf$ satisfies the second definition.
