This is a proof verification question. Here, $\, f$ is continuous and $\alpha$ is of bounded variation.
My only tools are the sums, for a given partition $P = \{a=x_0 < \ldots < x_n = b \}$ of $[a,b]$ : $$S_\alpha(f,P)= \sum_{j=1}^n f(t_j) [\alpha(x_j)-\alpha(x_{j-1})], \,t_j\in[x_{j-1},x_j]$$
and a criterion of integrability:
If $\alpha \in BV([a,b])$, and $f:[a,b] \to \mathbb{R}$ is continuous, then there is a unique number $I$ such that for every $\epsilon > 0$ there is a $\delta > 0$ such that when $P$ is a partition with $\Vert P \Vert<\delta$, $$\mid \, S_{\alpha}(f,P)-I\mid \,<\epsilon.$$
My proof: Since $f(x)\geq 0$ for all $x \in [a,b]$ and $\alpha$ is increasing, we have that $S_{\alpha}(f,P)\geq 0$ for any partition $P$.
Let $\epsilon >0$, then there is a $\delta >0$ such that if $\Vert P \Vert<\delta$, $\, \mid S_{\alpha}(f,P)-I\mid \,<\epsilon.$
Now assume $I<0$. Then $-I >0$, and so $$\epsilon > \,\mid S_{\alpha}(f,P)-I \mid \, > \,\mid S_{\alpha}(f,P)-0 \mid \geq 0,$$
so $I$ is not unique! Hence $\int_a^bf d\alpha = I \geq0.$
Is this proof correct?
Using only this facts, is there a direct proof?