In Øksendals Stochastic differential equations on page 269 he defines the bank proess as a process satisfying:
$$dX_0(t)=\rho(t,\omega)X_0(t)dt; X_0(0)=1.$$
He later says that the bank process satisfies
$$X_0(t)=\exp\left(\int_o^t\rho(s,\omega)ds\right).$$
But how does he get to that result? I know it resembles calculus, but in order for us to use ordinary calculus we would have to have that $dX_0(t)=\rho(t,\omega)X_0(t)dt$ is a differential equation, and that $\rho(t,\omega)$ is Riemann-integrable $\omega$-wise. When we say that $dX_0(t)=\rho(t,\omega)X_0(t)dt$ in terms of stochastic calculus we are not assuming anything about differentiability, we are just assuming that
$$X_0(t,\omega)=X_0(0)+\int_0^t\rho(s,\omega)X_0(s)ds.$$
And we don't assume that $\rho(t,\omega)$ is Riemann-integrable $\omega$-wise, we just assume that the $$P\left(\int_0^t |\rho(s,\omega)X_0(s)|ds<\infty\right)=1,$$
and that $\rho$ is progressiely measurable(Øksendal only assume measurable, not progressively, but I think it should be progressively to get adaptedness).
But how can we get to $X_0(t)=\exp\left(\int_o^t\rho(s,\omega)ds\right)$ by just using the measure-theoretic assumptions, and not the basic assumptions from calculus like differentiability and Riemann-integrability?
I think you can use Ito's lemma like you would with the other processes in the market:
$$dX_0(t)=\rho(t,\omega)X_0(t)dt+\sum_{j=1}^m0\cdot dW_j(t)$$
Consider $F(t,X_0)=\ln(X_0)$, then $$dF(t)=\frac{\partial F}{\partial x_0}dX_0(t)=\rho(t,\omega)dt$$ Therefore $$\ln(X_0(t))-\ln(X_0(0))=\int_0^t\rho(s,\omega)ds \implies X_0(t)=\exp\bigg(\int_0^t\rho(s,\omega)ds\bigg)$$