Summation of Supremum

Let $[a,b]$ be an interval in $\mathbb{R}$. Let $P=\{x_0,...,x_n\}$ be a partition of $[a,b]$ and $f$ be a real function bounded on $[a,b]$. Let $\alpha$ be a monotonically increasing function on $[a,b]$

Suppose $\forall s\in \Pi_{1≦i≦n} [x_{i-1},x_i], \sum_{i=1}^n f(s_i)[\alpha(x_i) - \alpha(x_{i-1})]< A$ for some $A\in \mathbb{R}$.

I don't know how to prove that $U(P,f,\alpha)≦A$.

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What does $U(\cdot,\cdot,\cdot)$ mean? –  John Bentin Nov 28 '12 at 11:49
@John Bentin : I would be temped to think that Katlus is studying the Riemann-Stieltjes integral (he as posted an other question on the subject today). And using the notations of Wikipedia: en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral –  Sebastien B Nov 28 '12 at 12:54
If I understand well your question $$U(P,f,\alpha)=\sum_{i=1}^n(\sup_{x\in [ x_{i-1} ,x_{i} ]} f(x))(\alpha(x_{i})-\alpha(x_{i-1}))\,.$$
Then for $i=1,\dots,n$ you can find sequences $(s_{i,j})_j$ in $[x_{i-1},x_i]$ such that $f(s_{i,j})\to \sup_{x\in [ x_{i-1} ,x_{i} ]} f(x)$
then $$\sum_{i=1}^n f(s_{i,j})[\alpha(x_i)-\alpha(x_{i-1})]\to U(P,f,\alpha)$$ when $j\to\infty$. Using your asumption on the sums on the left hand side terms you obtain
$$A\geq U(P,f,\alpha)\,.$$