Let $X(t)$ be a Brownian motion. I know that the integral \begin{equation} Y(t) = \int_0^t d\tau ~ X(\tau) \end{equation} is well-defined, since Brownian motion $X(\tau)$ is a.s. continuous. Thinking of $X$ as a continuous process, it makes thus intuitively sense that $Y(t)$ is a well-defined random variable.
On the other hand, I can think of the trajectory of $X(t)$ as a fractal. Then, I would expect that the path length between $X(t_1)$ and $X(t_0)$ for any $t_0 \neq t_1$ is infinite, and $Y(t)$ hence diverging. Why is this intuitive thinking wrong? Is Brownian motion not a fractal?
You are right that Brownian motion is a fractal and that the length of the path $[0,1]\ni t\to X(t)$ is infinite, almost surely. But the integral defining $Y(t)$ is not measuring path length, but rather the area of the region bounded by the graph of $X|_{[0,t]}$ and the $t$-axis (with area below the axis counted as negative, as usual).