Lemma of Gronwall Let $T \in \mathbb R^+\cup \{\infty\}$, $t_o \in [0,T)$, $a,b \in L^\infty(t_0,T)$ and $\lambda \in L^1(T_0,t)$, $\lambda(t) \geq 0$ for almost all $t \in (t_o,T)$. From the inequality $$a(t) \leq b(t) + \int_{t_0}^t\lambda(s)a(s)ds \,\,\,\,\,$$ a.e. in $(t_o,T)$ it follows  $$a(t) \leq b(t) + \int_{t_0}^t e^{\phi(t)-\phi(s)}\lambda(s)b(s) \,ds$$ for almost all $t\in (t_0,T)$ where $\phi(s):=\int_{t_o}^s \lambda(\tau)\,d\tau$.
Is there a counterexample for the case that $\lambda$ is negative? And what changes about the implication if $t<t_o$?
 A: Here's a simple example of what happens when $\lambda$ is not non-negative.
Take $t_0 =0$, $T=\infty$, and  $b(t) =1$ and $\lambda(t) =-1$ for all $t \ge 0$.  Further assume we have the equality
$$
a(t) = b(t) + \int_0^t \lambda(s) a(s) ds = 1 - \int_0^t a(s) ds
$$
for $t \ge 0$.  Write $F(t) = \int_0^t a(s) ds$.  Then the equality reads
$$
F'(t) = 1 - F(t) \Rightarrow (e^t F(t))' = e^t \Rightarrow F(t) = 1- e^{-t}.
$$
Plugging back in shows that $a(t) = e^{-t}$.
Now let's consider the RHS of the Gronwall inequality.  We compute
$$
b(t) + \int_0^t e^{\phi(t)-\phi(s)} \lambda(s) b(s) ds = 1 - e^t \int_0^t e^{-s} ds = 2 - e^{t}.
$$
The inequality then holds if and only if 
$$
e^{-t} = a(t) \le b(t) + \int_0^t e^{\phi(t)-\phi(s)} \lambda(s) b(s) ds = 2- e^{t}
$$
for all $t \ge 0$, which is equivalent to
$$
\cosh(t) \le 1 \text{ for all }t \ge 0,
$$
a contradiction.
For your second question I can't really give a good answer, as it's not clear to me what you want to assume.  Are we supposed to assume that the first inequality also holds for $t < t_0$?  If not, then we have no information about $a$ before $t_0$, so how could we say anything about the function?
