solution of an integral equation in measurable functions Let $\phi(t)$ be a positive continuous function on $[0,\infty)$ and $f(t,x)$ be a continuous function of two variables such that 
$$
|f(t,x)|\leq \phi(t)|x|.
$$
Suppose $\int_0^\infty\phi(t)<\infty$.  If a function $y$ satisfies for all $t\geq 0$, 
$$
|y(t)|\leq \int_0^t|f(s,y(s))|ds,
$$
prove that $y(t)=0$ for all $t$. How to prove?
 A: If we can assume $y(t)$ is continuous, we can "Gronwall it"; by which I mean:
If
$|y(t)|\leq \int_0^t|f(s,y(s))|ds, \tag{1}$
with
$|f(t,x)|\leq \phi(t)|x|, \tag{2}$
then we conclude that
$|y(t)|\leq \int_0^t\phi(s) \vert y(s) \vert ds, \tag{3}$
to which may apply the integral form of Gronwall's inequality.  For the present purposes, it may be taken to assert that, for continuous functions $u(t)$ and $\beta(t) \ge 0$ on an interval $I \subset \Bbb R$ and constant $\alpha \ge 0$ with
$u(t) \le \alpha + \int_{t_0}^t \beta(s) u(s) ds \tag{4}$
for $t_0, t \in I$, $t_0 \le t$, we have
$u(t) \le \alpha e^{\int_{t_0}^t \beta(s) ds}. \tag{5}$
Taking $\vert y(t) \vert = u(t)$ and $\beta(t) = \phi(t)$ we see that the hypothesis of Gronwall's are met in the present case.  Since $\alpha = 0$ we conclude $\vert y(t) \vert = 0$, that is, $y(t) = 0$ for $t \ge 0$.   QED.
P.S.  There is another form of Grownall's inequality which applies to certain measurable functions and may yield a stronger result for the present problem, but I'm too tired right now to write that version of my argument up.  The interested reader might want to check out the linked citing and take it from there.  End:  Post Script.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Here is a cute proof: We immediately get $|y(t)| \leq \int_0^t \phi(s)|y(s)|ds$ for all $t \geq 0$. 
Now define the function: 
$g(t) = (\int_0^t\phi(s)|y(s)|ds)e^{-\int_0^t\phi(s)ds}$. 
We see that:
$g(0)=0$
$g(t)\geq 0$ for all $t \geq 0$
$g'(t)\leq 0$ for all $t \geq 0$.  
So $g(t)=0$ for all $t \geq 0$.
