Let $A_t:=\int_0^tf(B_s)ds,\quad t\geq0$

with $f$ continuous and $B$ standard Brownian motion.

What is the correct argument that $A$ is of finite variation? Because it can be written as an integral?

And why is $\int_0^t f(B_s)dB_s$ a local martingale? Because $B$ is a local martingale and $f$ is continous?

  • $\begingroup$ If $f \equiv 1$ then $A_t=t$ is not of bounded variation on $[0,\infty)$. $\endgroup$ Mar 12, 2021 at 9:52
  • $\begingroup$ @KaviRamaMurthy I meant finite variation. Does this make a difference? $\endgroup$
    – user826130
    Mar 12, 2021 at 9:56
  • $\begingroup$ As far as my definitions go, there is no difference. $\endgroup$ Mar 12, 2021 at 10:01

1 Answer 1


Since $B_s$ has continuous paths, $f(B_s)$ has continuous paths and hence, $A_t$ has $C^1$ paths (that's the fundamental theorem of calculus). However, any $C^1$ process has locally bounded variation (some authors omit this word) since any $C^1$-function has locally bounded variation.

As for why $\int_0^t f(B_s)\textrm{d}Bs$ is a local martingale, that simply follows from the construction of the Itô integral.


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