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For the sake of example, suppose we are on the real line and that $H(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac Delta function. According to the theory of distributional derivatives, one has


in the sense of distribution, so that $\int \phi(x) \delta(x)dx = -\int \phi'(x) H(x)dx$ for all compactly supported nice functions $\phi$.

My question is whether the same definition carries over to Riemann-Stieltjes integration. In particular, for a cumulative distribution function $F$, I am trying to evaluate $\int \phi(x)H'(x)dF$ but, does this expression equal $\int \phi(x)\delta(x)dF=\phi(0)dF(0)$?.

My suspicion is that the distributional derivative here must take into account the $dF$ term, so integration by parts of $\int \phi(x)H'(x)dF$ does not give the usual expression. If $F$ has a density $f(x)$, then one gets $\int \phi(x)H'(x)f(x)dx=-\int (\phi f)' H(x)dx=\phi(0)f(0)$ which is as expected, so what happens when $F$ does NOT have a density function $f$? In particular how does this generalize to arbitrary distributions?

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Try it for $F(x) = H(x)$. The square of the Dirac Delta does not make sense. – Robert Israel Jun 8 '12 at 21:12
@Robert Israel: I agree. But does that mean there is no way to define the distributional derivative under that particular Stieltjes integral – Alex R. Jun 8 '12 at 22:48

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