How has the author obtained Riemann-Stieltjes sum of $g$ wrt $f$ as $f(b)g(b) - f(a)g(a) - \sum_{k=1}^{2n} f(x_i) [g(y_k) - g(y_{k-1})]$

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In the book of The elements of Real Analysis by Bartle, at page 232, it is given that

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Let $J = [a,b]$, and $P$ be a partition of $J$. Define $$S(P; f, g) = \sum_{k=1}^n g(e_k) [f(x_k)- f(x_{k-1})],$$ where $x_{k-1} \leq e_k \leq x_k$, and $x_i$s are the end points of $P$. Let $Q= (y_0, y_1, ..., y_{2n})$ be partition of $J$ obtained by using both $e_k$ and $x_k$ as partition points; hence $x_{2k} = x_k$ and $y_{2k-1} = e_k$. Add and substract the terms $f(x_{2k}) g(x_{2k})$, $k=0,...,n$ to $S(P; f, g)$, and rearrange to obtaion $$S(P; g, f) = f(b)g(b) - f(a)g(a) - \sum_{k=1}^{2n} f(x_i) [g(y_k) - g(y_{k-1})],$$ [...]

we get $$\int_a^b f dg + \int_a^b gdf = f(b)g(b) - f(a)g(a)$$

My question is that how the author did obtain $$f(b)g(b) - f(a)g(a) - \sum_{k=1}^{2n} f(x_i) [g(y_k) - g(y_{k-1})]$$ from adding the terms $f(x_{2k}) g(x_{2k})$, $k=0,...,n$ to $$\sum_{k=1}^n g(e_k) [f(x_k)- f(x_{k-1})].$$