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What is the geometrical interpretation of Riemann Stieltjes Integration ?

We know that for Riemaan integration $\int_{a}^b f(x)dx$ represents the area bounded by the curve $y=f(x)$& the straight lines $x=a$ & $x=b$.

But when we integrate $f(x)$ with respect to another function $g(x)$ then which area represents that integration geometrically ?

Empty
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2 Answers2

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Please read the paper entitled

"A Geometric Interpretation of the Riemann-Stieltjes Integral" American Mathematical Monthly archive
Volume 95 Issue 5, May 1988 Pages 448-455

Ali
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Imagine that instead of plotting the graphic of $f(x)$ "based" on the straight monodimensional line Ox, rising out of it along a second dimension Oy, you would instead plot it "based" on the possibly curvy bidimensional line $g(x)\in xOy$, rising out of it along a third dimension Oz, and forming a straight angle with the horizontal plane xOy.

Lucian
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  • For instance, if g would be a $($semi$)$circle, and f a constant, you'd get a cylindrical shape. – Lucian Sep 30 '14 at 03:36
  • It is not clear to me.. – Empty Sep 30 '14 at 16:44
  • @SayantanPanja: Take a sheet of paper, and draw on it the two axes of coordinates, and the graphic of a random function $g(x)$. Then imagine that from each point of the bidimensional plot of $g(x)$, a beam of light comes out of the sheet of paper, whose direction is perpendicular on that sheet of paper, and whose length is equal to the value of $f(x)$ in that point, $z=f(y)$, where $y=g(x)$. If the value is positive, then the beam of light goes upwards from the sheet; if not, then it goes downwards. You'll get a wavy surface, whose signed area is $\displaystyle\int fdg$. – Lucian Sep 30 '14 at 16:58