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If $f$ and $g$ are both Riemann-Stieltjes Integrable with respect to a monotonic function $\alpha$, is it ture that $f(g(x))$ is still integrable with respect to $\alpha$? Thanks for your attention.

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No. Take $\alpha(x) = 1/x^3$ and $f(x) = g(x) = x^2$. we have $$ \int_1^\infty f(x) d\alpha(x) = \int_1^\infty g(x) d\alpha(x) = -3\int_1^\infty \frac 1 {x^2} dx = -3 $$ but $$ \int_1^\infty f(g(x)) d\alpha(x) = -3\int_1^\infty dx = -\infty $$

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Thank you, but when the integral is defined on an closed interval $[a,b]$, does this kind of function still exist? –  Golbez Oct 18 '12 at 13:37
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@Golbez There are. Take $\alpha(x) = x^{3/2}$, $f(x) = x^2$, $g(x) = 1/x$ on the interval $[0, 1]$. –  AlbertH Oct 18 '12 at 13:59
    
@Albert the second function you defined f(x) = x^2 & g(x) = x^-1 might not work out. fog(x) has only one point of discontinuity. any function with finite points of discontinuity in Interval [a,b] is Reimann Integrable provided α(x) is continuous at those points. –  user52514 Dec 10 '12 at 17:45

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