Easy question on stochastic integral

Question

Is it legit to say that $$\int_0^tdB_s=B_t-B_0$$ I'm a little bit confused because I know that the Brownian motion is not differentiable, but doing so we are treating the $dB_s$ as it is.

Could someone help me to understand this step?

Answer 1

The answer is: by definition of the integral.

In order to define the integral with respect to the Brownian motion for general integrands, one first define the (pathwise) integral for integrands that are simple process as

$$\int _0^t g(u)dB_u := \sum_{i=0}^{k-1}= \phi_i (B_{t_{i+1}}-B_{t_i}), \tag*{(*)}$$

where $g$ is a simple process of the form $$g(t) = \phi_0 I_{[t_0,t_1]}(t)+ \sum_{i=0}^{k-1} \phi_i (I_{t_{i+1}}-I_{t_i}).$$

In your case $g(t)$ is the simplest of the simple processes. We have just $g(t)=1,$ and substituting into (*), we get your result.