Given $a $$\alpha(x)=\begin{cases}
\alpha(a),&\text{if }a\le x Let $f:[a,b]\to\Bbb R$ continuous. Prove that $f\in R(\alpha)$ and that $$\int_a^bf(x) d\alpha(x)=f(c)\big[\alpha(b)-\alpha(a)\big]$$ My attempt Since $f$ is continuous on $[a, b]$ then $ f \in R (\alpha) $ on $[a,b]$ Then for a partition $P$ such that $a=x_{0} We have $U(P,f,\alpha)=M_{i}\big(\alpha(b)-\alpha(a)\big)$ and $L(P,f,\alpha)=m_{i}\big(\alpha(b)-\alpha(a)\big)$ Can anyone help me to conclude the result please? Thanks for your help
Definite integral against a weight function
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real-analysis
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1I have edited in a more informative title. I'm not certain I've captured the essence of the question. I encourage improvements. – 2012-12-10
2 Answers
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This seems to be a Riemann-Stieltjes integral, so I would approach the problem in the following way:
As you mention, since $f$ is continuous, $f\in R(\alpha)$, thus, for a partition $\Gamma=\{a=x_0
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Hint: So $\alpha$ is a step function that takes on two values and is discontinuous at $c$. This tells you that given a partition $P=\{x_1,\ldots,x_n\}$, most of the terms $\alpha(x_i)-\alpha(x_{i-1})=0$. The only time this doesn't happen is near $c$. This means the upper and lower sums will be very simple to write.
There are two cases: (1) $c \in P$, say $x_k=c$ for some $k$, and (2) $c \notin P$, say $x_{k-1}