Path of diffusion process with discontinuous drift Let $(B_t)$ be a standard Brownian motion on some probability space and let $X_t$ be the process defined by the SDE $dX_t = \mu_t dt + dB_t$, where $\mu_t$ is adapted, deterministic, and only takes the values $0$ or $1$.  (For example, $\mu_t = 0$ on $(2n, 2n+1]$ for all $n \geq 0$ and $1$ everywhere else.)
Does it follow that $X_t$ has continuous sample paths?  (Under the same probability measure that makes $(B_t)$ a standard BM?)
I think it does -- use the Girsanov theorem to get a new probability measure so that with $X_t$ is now a standard BM.  Because the new probability measure is mutually absolutely continuous with the old one, it should follow that $X_t$ is also continuous almost everywhere.
Am I right with this line of thought, or am I missing something?

Applying Girsanov's theorem is overkill. Note that, by definition,
$$X_t = X_0 + \int_0^t \mu_s \, ds+ B_t, \qquad t \geq 0.$$
We know that $t \mapsto B_t$ is continuous (almost surely); moreover, it is well-known that mappings of the form
$$t \mapsto I(t) := \int_0^t \mu_s \, ds$$
are continuous  whenever the integral is well-defined. (Just consider e.g. $\mu(s) = 1_{[1,2]}(s)$; draw a picture to see that $t \mapsto I(t)$ is continuous, it doesn't have any jumps.)
This means that $t \mapsto X_t$ is continuous almost surely since it is the sum of the continuous functions.
