Almost sure convergence of stochastic process Suppose that we have a (almost surely) continuous stochastic process
$\{ X_{t} \}_{t \geq 0}$ on $[0,1]$ with non-stochastic initial value
$X_{0} = x_{0} \in [0,1]$ and exponentially decreasing expectation
$E(X_{t}) = x_{0} e^{-\gamma t}$, where $\gamma > 0$.
For the corresponding discrete-time process $\{ X_{n} \}_{n =
  0}^{\infty}$, an application of the Markov inequality and the
Borel–Cantelli lemma shows that $\lim_{n \rightarrow \infty} X_{n} =
0$ almost surely. Is the same true for the continuous-time process
$\{ X_{t} \}$, i.e. do we have $\lim_{t \rightarrow \infty} X_{t} = 0$
almost surely?
Originally, the process $\{ X_{t} \}$ is the (almost surely) unique,
continuous, and Markovian solution of the Ito stochastic differential
equation
\begin{equation*}
dX_{t} = -\gamma X_{t} dt + k \sqrt{\gamma} \sqrt{X_{t}^{3}(1-X_{t})}dW_{t}, \ X_{0} = x_{0} \in [0,1]
\end{equation*}
where $k > 0$ and $W_{t}$ is a Brownian motion. Does this ensure the almost sure convergence towards $0$?
 A: By applying Ito's formulq we get $e^{\gamma t}X_t$ is a local martingale, so it is a supermartingale for it is positive, and so it is for $X_t$. Then by some classical argument(for example an exercise in Chapter 1 of Brownian motion and stochastic calculus) we know any positive supermartingale converge almost surely. Denote its limit by $X_{\infty}\geq 0$, Fatou's lemma gives
$$E[X_{\infty}]\leq\liminf_{t\rightarrow\infty}E[X_t]=0$$
which allows to show $X_{\infty}=0$ a.s.
A: @Higgs88 Thank you very much for taking the time to answer this
question. I have decided to write out the details myself to see if I
understand it right.
We want to prove that if $\{ X_{t} \}_{t \geq 0}$ is a continuous, supermartingale with asymptotically vanishing expectation
$\lim_{t \rightarrow \infty} E(X_{t}) = 0$, then $\lim_{t \rightarrow \infty }
X_{t} = 0$ almost surely. Following @Higgs88, by Doob's first
martingale theorem, $\{ X_{t} \}_{t \geq 0}$ has a limit $X_{\infty}
\in [0, \infty[$ almost surely. In particular, we have almost sure
convergence in discrete time, i.e. $\lim_{n \rightarrow \infty} X_{n}
= X_{\infty}$ almost surely, and thus, by Fatou's lemma applied to the non-negative sequence $\{ X_{n} \}$, we get 
\begin{equation}
E(X_{\infty}) = E( \lim_{n \rightarrow \infty} X_{n} ) = E( \liminf_{n \rightarrow \infty} X_{n} ) \leq \liminf_{n \rightarrow \infty}
E(X_{n}) = \lim_{n \rightarrow \infty} E(X_{n})= 0
\end{equation}
But by the Markov inequality, $P(X_{\infty} \geq
\epsilon ) \leq \frac{1}{\epsilon} E(X_{\infty}) = 0$ for all
$\epsilon > 0$, and $X_{\infty} = 0$ almost surely.
To see how to apply the above result to the Ito solution $\{ X_{t}
\}$, we need to show that it is a supermartingale. However, put $Y_{t}
= e^{\gamma t} X_{t}$ and notice that $dY_{t} = k \sqrt{\gamma}
\sqrt{e^{-3\gamma t} Y_{t}^{3} (1-e^{-\gamma t} Y_{t})}dW_{t}$ by
Ito's lemma, and $\{ Y_{t} \}$ is a non-negative local martingale and
thus a supermartingale. Consequently, $\{ X_{t}\}$ is a
supermartingale.
However, I wonder if some of the assumptions might be weakened or removed,
in particular the supermartingale assumption.
