Let $B$ be a Brownian motion and $M_t = \int_0^t \mathbb{1}_{B_s=0} dB_s$. It can be shown that $(M_t)_{t \geq 0}$ is a local martingale. Now, I want to calculate the quadratic variation of $M$.
In general, if $I_t = \int_0^t f(s) dB_s$ is an Itô-integral, then the quadratic variation of $I_t$ is
\begin{align}
\langle I \rangle_t = \int_0^t f^2(s) ds.
\end{align}
So,
\begin{align}
\langle M \rangle_t = \int_0^t \mathbb{1}_{B_s=0}^2 ds = \int_0^t \mathbb{1}_{B_s=0} ds.
\end{align}
However, I do not see directly why $ \int_0^t \mathbb{1}_{B_s=0} ds = 0$?
Best Answer
We have $\mathbb{P}(B_s=0)=0$ for all $s>0$, and therefore it follows directly from Tonelli's theorem that
$$\mathbb{E} \left( \int_0^t 1_{\{B_s=0\}} \, ds \right) = \int_0^t \underbrace{\mathbb{E}(1_{\{B_s=0\}})}_{=\mathbb{P}(B_s=0)} \, ds = 0.$$
Since the integrand is non-negative this implies
$$\int_0^t 1_{\{B_s=0\}} \, ds = 0 \quad \text{a.s.}$$