# Questions regarding finite variation and local martingale

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?

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

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.