The following fact is stated in my lecture notes: Let $f: \mathbb{R}_+ \to \mathbb{R}_+$ be an increasing and continuous function satisfying $f(0) = 0$. Then there exists a continuous real-valued martingale $M$ with quadratic covariation process $[M]$ given by f, i.e $[M] = f$. (For clarification: We define the quadratic covariation process of a continuous local martingale $M$ as the unique process , denoted by $[M]$, such that it starts in $0$, has locally finite variation and makes the process $M^2 - [M]$ a continuous local martingale.)
I am not sure how to construct such a martingale. I already tried the follwowing:
Given a continuous local martingale $M$ and a progressive process $V$ (satisfying some conditions), I know that the covariation process of the stochastic integral $V \cdot M$ is given by $$ [V \cdot M]_t = \int_0^t V^2_sd[M]_s $$ and this expression must somehow equal $f(t)$ for all $t$. Let us try Brownian motion as $M$. We get (since the covariation of brownian motion is just $t$): $$ [V \cdot B]_t = \int_0^tV^2_s ds. $$ So it seems that I can set $V_s := \sqrt{\frac{df}{dt}(s)}$ and get with $M:= V\cdot M$ $$ [M]_t = [V \cdot B]_t = \int_0^t (\sqrt{\frac{df}{dt}(s)})^2 ds = f(t) - f(0) = f(t), $$ as I wanted.
However, I cannot assume that $f$ is differentiable (or: the measure $\mu_f$ induced by $f$ does not necessarily satisfy $\mu_f << \lambda$). So my argument does not work.
After reading this, I tried to use the Lebesgue decomposition of the measure induced by $f$, i.e. decompose $f(t) = \int g(s) d\lambda(s) + h(s)$, where the integral represents the absolutely continuous part and $h$ the singular component. Using Brownian motion, I can deal with the integral part, but I do not know how to deal with the singular part. So, assuming that $f$ is purely singular (e.g. the Cantor function), I am nowhere.
I don't know where I can start to tackle the problem concerning the singular part of $f$ and I would be glad if somehow could hint me into the right direction.
Thanks in advance!
