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Given $a , define $\alpha:[a,b]\to \Bbb R$ as follows: $\alpha(a)<\alpha(b)$, $\alpha(c)\in \big[\alpha(a),\alpha(b)\big]$, and also

$\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

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    I 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 2

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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, we have $\int_a^b fd\alpha=\lim_{n\to\infty}\sum_{i=1}^n M_i[\alpha(x_i)-\alpha(x_{i-1})].$ Now, since we're told the only place where $\alpha(x_i)-\alpha(x_{i-1})\neq0$ is when $x_i=c$ or $x_{i-1}=c$, we're left with $\int_a^bfd\alpha=M_{k}(\alpha(c)-\alpha(x_{k-1}))+M_{k+1}(\alpha(x_{k+1})-\alpha(c)).$ Recalling that for $x we have $\alpha(x)=\alpha(a)$ and similarly for $\alpha(b)$, we have $M_k\alpha(b)-M_{k+1}\alpha(a)+\alpha(c)(M_{k+1}-M_k).$ Now, by the continuity of $f$, we have the $M_k$ and $M_{k+1}\to f(c)$ which implies the second term converges to $0$. Thus we have the result.

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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} for some $k$. Find the upper and lower sums in each case, then compare their respective infimum and supremum.