What does it mean that $X_T$ is a solution to a Brownian Motion driven SDE ($dX_t = ....$)? Please consider the
$$
X_t = X_0 + \int^t_of(X_s,s)ds+\int^t_og(X_s,s)dB_s
$$ where $B_t$ is Brownian Motion. This can also be expressed as:
$$dX_t=f()dt+g()dB_t$$
What does it mean that $X_T$ is a solution to the above stochastic differential equation?
I can't seem to intuitive explain this. The way I will put it is:
For a given Brownian motion realization it $X_T$ is a solution to the SDE above.. Is this correct or is there more it?
 A: There's actually two definitions of solution to SDE. Strong and weak. 

Strong solution. Given a probability space $
(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space adapted to $\mathcal F_t$, a solution to $dX=fdt+g~dB(t)$ with initial condition $x\in \mathbb R$ is a continuous process adapted to $\mathcal F_t$ such that $X(t)=x+\int_0^t fdt+\int_0^t g~dB(t)$.

There is also

Weak solution. A process $X(t)$ is called a weak solution to $dX=fdt+g~dB(t)$ if there exists a probability space $
(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space adapted to $\mathcal F_t$ such that $X(t)=x+\int_0^t fdt+\int_0^t g~dB(t)$.

Note that strong solution implies weak solution.
As an example of something that has weak but not strong solution, consider Tanaka's equation
$$dX(t)=\text{sign}(X(t))dB(t)$$
It can be shown that this has no strong solution by considering local times. However let $X(t)$ be a Brownian motion and then $X(t)$ is a weak solution. Note that $\int_0^t\text{sign}^2(s)ds=t<\infty$ so the Ito integral $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\int_0^t \text{sign}(s) dX(s)$ exists. Noting that the quadratic variation of this integral is $t$, and Ito integrals are continuous martingales, so we know $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\tilde{B}(t)$ is a Brownian motion. 
Thus, $dX(t)=\text{sign}(t)d\tilde{B}(t)$.
The difference here is that the Brownian motion is constructed in terms of our solution not the other way around.
