I know that this function [itex] f : [-2,2] \to \mathbb{R} [/itex](adsbygoogle = window.adsbygoogle || []).push({});

[itex] f(x) = \begin{cases} 1 & \textrm{ if } x \geq 0 \\ 0 & \textrm{ if } x < 0 \end{cases} [/itex]

is not Riemann-Stieltjes integrable with itself (that is, taking [itex] g = f [/itex] then [itex] f \not\in R(g) [/itex])

That is because both share a point of discontinuity, namely 0.

What I don't know is how do I justify this? for wich partition does the limit of the superior sums and the inferior sums don't equal? Shouldn't this integral be equal to 1? (all the terms in the sum cancel except at zero, where [itex] \triangle g = 1-0 = 1 [/itex], and [itex] f(0) = 1[/itex]

I think I'm not seeing it correctly.

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# Non Stieltjes integrable function

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