probability stat we consider the differential equation 
$$y'(t)=f(t,y(t)),\quad t \geq 0, y(0)=y_0$$
I haven't any idea, help me please.
 A: Hint: The condition $\sup_{t\in[0,T[}|y(t)|\le M$ tells you that either the limit exists, or $y(t)$ oscillates really fast when approaching $T$. The Lipschitz condition tells you that $f$ cannot go from very positive to very negative too fast, and thus prevents fast oscillations.
A: We need to further assume that $f$ is continuous.
Let
$$
\tilde M=\max_{(t,x)\in [0,T]\times [-M,M]} \lvert f(t,x)\rvert.
$$
Such $\tilde M<\infty$ exists due to continuity of $f$ in the compact $K=[0,T]\times [-M,M]$.
Let now $\{t_n\}_{n\in\mathbb N}\subset [0,T)$, with $t_n\to T$. Since the sequence $\{y(t_n)\}_{n\in\mathbb N}$ is bounded then it has a convergent subsequence $\{y(t_{n_k})\}_{k\in\mathbb N}$, with $y(t_{n_k})\to \xi$.
Then for every $t\in [0,T)$ we have that
$$
y(t)-y(t_{n_k})=\int_{t_{n_k}}^t f\big(s,y(s)\big)\,ds, \tag{1}
$$
for every $t\in [0,T)$, and hence
$$
\big\lvert y(t)-y(t_{n_k})\big\rvert \le \int_t^{t_{n_k}} \big| \,f\big(s,y(s)\big)\big|\,ds\le \tilde M\lvert t-t_{n_k}\rvert. \tag{2}
$$
Letting $k\to\infty$ in $(2)$ we obtain 
$$
\big\lvert y(t)-\xi\big\rvert \le \tilde M\lvert t-T\rvert,
$$
which implies that
$$
\lim_{t\to T^-}y(t)=\xi.
$$
Extending now continuously $y$ to $[0,T]$, as $y(T)=\xi$, we obtain from $(1)$ that
$$
y(t)-y(T)=\int_T^t f\big(s,y(s)\big)\,ds,
$$
and hence $y'(T)=f\big(T,y(T)\big)$, which means that $y$ satisfies the IVP in $[0,T]$.
For the second question, according to Picard-Lindelof's Theorem, there exists a solution $\tilde y$ of the initial value problem
$$
x'=f(t,x),\quad y(T)=\xi, \tag{3}
$$
defined in an interval $(T-\tau,T+\tau)$. Lipschitz condition guarantees uniqueness, and as the previous solution $y$ satisfies $(3)$ in $[0,T]$, then $y$ and $\tilde y$ coincide in their common domain, and hence $\tilde y$ extends $y$ to a solution in $[0,T+\tau)$, and therefore $y$ extends, and its not a maximally defined solution.   
