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For our purposes here, a singular function is a continuous function such that the part which is absolutely continuous with respect to Lebesgue measure is zero. For example, the Cantor function or "devil's staircase" is one such function.

This is a very soft question: I'm looking for situations where these functions come up in applied problems.

I do have one example already, which may give an idea of what I'm after. Consider defining a diffusion process on a domain with reflecting boundary. Physically this arises when discussing diffusion in a domain with insulating boundary. Roughly speaking this is a diffusion process which has some noise component, possibly some drift component inside the domain, and a delta function drift at the boundary, which instantly forces the process back inside the domain whenever it hits the boundary.

If we consider the contribution of the reflecting force on the boundary, we find that for almost every sample path, the contribution is a singular function. This is because the underlying process without the boundary force actually spends zero time on the boundary itself, which means the boundary force must have infinite intensity while it is applied. Yet at the same time the trajectories are continuous, so this contribution is continuous as well. Hence the overall contribution is singular.

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Well the maximum of a brownian motion would be a singular function (and it's very useful).

EDIT: you can also consider the local time $L_t$ (second EDIT: which you mentioned in your question), which gives you the Tanaka formula:

$$|W_t| = \int_0^t sign(W_s)\,dW_s + L_t$$

It can be used to extend Ito's formula to convex functions (I think a simplified assumption). So in finance you can use that to get a closed form formula for the price of a call.

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  • $\begingroup$ Good example, thanks for this one. $\endgroup$ – Ian Dec 12 '14 at 15:12
  • $\begingroup$ The Tanaka formula is basically my example in 1D when the boundary is $\{ 0 \}$. Still useful, though. $\endgroup$ – Ian Dec 12 '14 at 15:31
  • $\begingroup$ Ok. I've modified my question. Sorry it was a bit confusing. It does indeed sound like a Doob Meyer decomposition. My bad :p $\endgroup$ – user3371583 Dec 12 '14 at 15:39

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