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

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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?2012-10-18
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    @Golbez There are. Take $\alpha(x) = x^{3/2}$, $f(x) = x^2$, $g(x) = 1/x$ on the interval $[0, 1]$.2012-10-18
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    @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.2012-12-10