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As written in Wikipedia, the Dirac delta function is the derivative of the Heaviside function $${\displaystyle \delta (x)={\frac {d}{dx}}H(x)}$$ Hence the Heaviside function can be considered to be the integral of the Dirac delta function$$ {\displaystyle H(x):=\int _{-\infty }^{x}\delta (s)\,ds}$$ In this context, the Heaviside function is the cumulative distribution function of a random variable which is almost surely $0$.

I would like to show that the Heaviside function is a càdlàg adapted process of finite variation. Could you please help me?

Lely
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    Using the definition of variation can you show directly that the variation of $H$ is zero on every interval in $(-\infty,0)$ and one on every larger interval ? In other words: the variation of $H$ is $H$ itself. See also this. – Kurt G. Aug 09 '22 at 14:50
  • adapted to what? – Tobsn Aug 10 '22 at 16:18
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    It will be adapted to every filtration since it is a deterministic function. The cadlag and finite variation property can easily be seen by directly checking "the" defnition of both. For this, it would be better to simply think of H as $H(x) = 0$ for $x \leq 0$ and $H(x) = 1$ for $x > 0$. – unwissen Aug 10 '22 at 21:11
  • Thank you, your comments answer my question! – Lely Aug 11 '22 at 08:06

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