Does this type of ODE satisfies unicity? ( right hand side f(t,y) = g(t) y, with g in L1 ) I have the following ordinary differential equation on $(0,1)$
$$
\dot y(t) = g(t) y(t) \quad \text{ almost everywhere in } (0,1)
$$
Where $g(t)$ only satisfies being integrable ($\int_0^1 |g(x)|dx < \infty $) and we are looking for absolutely continuous solutions $y(t)$ on $[0,1]$. (so, as a consequence, the solutions we look for are bounded).
The standard Picard's existence and unicity theorem requires $g(t)$ to be continuous, which is not the case in here. Can I ensure uniqueness of solution?
It is easy to see that the zero function is a solution, similarly, if I have a solution and an interval $[t_0,t]$ such that $y(s) \neq 0$, has constant sign for all $s \in [t_0,t]$, then it is also easy to see that
$$
y(s) = y(t_0) e^{\int_{t_0}^s g(x)dx }. 
$$
With this formula, I can extend such solution over the whole interval $(0,1)$ and notice that it would never touch the $0$, since, for that to happen, the integral above the exponential has to value $-\infty$; which doesn't happens because $g$ is integrable.
But I'm unsure on how to claim unicity of solution, as in, for some misterious $g$, a wild solution could emerge from 0. 
 A: Let us begin with a lemma: if $g \in L^1([0,1])$ and $t \in [0,1]$ then, for every $n \ge 1$,
$$\int _0 ^t \int _0 ^{s_1} \dots \int _0 ^{s_{n-1}} g(s_1) \ g(s_2) \dots g(s_n) \ \mathrm d s_n \dots \mathrm d s_2 \ \mathrm d s_1 = \frac 1 {n!} \left( \int _0 ^t g(s) \ \mathrm d s \right) ^n \ .$$
The proof is entirely analogous to the one here; its core idea is that the cube $[0,t]^n$ (which is the domain of integration of the right-hand side) is the union of $n!$ simplices identical to the one described by $\{(s_1, \dots, s_n) \in \mathbb R^n \mid 0 \le s_n \le \dots \le s_1 \le t\}$ (which is the domain of integration of the left-hand side), and these simplices overlap only on the boundaries. Furthermore, the factors in $g(s_1) \dots g(s_n)$ all commute.
Now, on to the main result: we want to prove that if $y(0) = 0$ and $y$ is absolutely continuous, then $y=0$ on $[0,1]$. To this end, notice that integrating the differential equation gives $y(t) = \int _0 ^t g(s_1) y(s_1) \ \mathrm d s_1$. Replacing the inner $y(s_1)$ by its corresponding integral, one gets $y(t) = \int _0 ^t \int _0 ^{s_1} g(s_1) \ g(s_2) y(s_2) \ \mathrm d s_2 \ \mathrm d s_1$, and inductively
$$y(t) = \int _0 ^t \int _0 ^{s_1} \dots \int _0 ^{s_{n-1}} g(s_1) \ g(s_2) \dots g(s_n) \ y(s_n) \ \mathrm d s_n \dots \mathrm d s_2 \ \mathrm d s_1 \ .$$
Making now repetead use of the usual inequality $| \int _a ^b f | \le \int _a ^b |f|$ we bring the modulus function inside all the integrals, getting
$$|y(t)| \le \int _0 ^t \int _0 ^{s_1} \dots \int _0 ^{s_{n-1}} | g(s_1) \ g(s_2) \dots g(s_n) \ y(s_n) | \ \mathrm d s_n \dots \mathrm d s_2 \ \mathrm d s_1 \le \dots$$
Since $y$ is continuous on the compact interval $[0,1]$, $|y|$ will have a finite upper bound $M$, which allows us to continue the above (also using the lemma):
$$\dots \le M \int _0 ^t \int _0 ^{s_1} \dots \int _0 ^{s_{n-1}} | g(s_1) \ g(s_2) \dots g(s_n) | \ \mathrm d s_n \dots \mathrm d s_2 \ \mathrm d s_1 \le M \frac 1 {n!} \left( \int _0 ^t |g(s)| \ \mathrm d s \right) ^n \ .$$
To conclude, if $I = \int _0 ^1 |g(s)| \ \mathrm d s$, we may deduce the inequality
$$|y(t)| \le M \frac {I^n} {n!} \quad \forall n \ge 1 \ .$$
But this is great, because the right-hand side converges to $0$, which means that $y(t) = 0$ for all $t \in [0,1]$, which proves uniqueness!
Just in case that you do not know why $\lim _n \frac {I^n} {n!} = 0$ when $I \ge 0$, notice that if $x_n = \frac {I^n} {n!}$ then $x_{n+1} = \frac x {n+1} x_n$, which shows that the sequence $(x_n) _{n \ge x-1}$ decreases. Since $x_n \ge 0$ for all $n$, we deduce that $(x_n)$ has a finite limit $L \ge 0$. Passing to the limit in the recurrence relation gives $L = 0 \cdot L$, whence $L=0$.
