I am researching the existence and uniqueness of the global solution of the original differential equation (ODE) of the form: $$\dot{x}(t) = f(t,x), \ \ t \geq 0 ; \ \ x(0) = x_0. \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
There are many results that have been given, for example:
$\bullet$ In Khalil's book [1, Theorem 3.2], the assumptions made are $f(\cdot,x)$ is a piecewise continuous function for any given x and $f(t,\cdot)$ is a Lipschitz continuous function for all t then (1) has a global absolutely continuous (AC) solution on $[0, +\infty)$.
$\bullet$ In Sontag's book [2, Theorem 54, Proposition C.3.8], it says that if $f(\cdot,x)$ is a measurable function for any given x, and $\|f(t,x_0)\|$ is bounded for all t, and $f(t,\cdot)$ is a Lipschitz continuous function for all t then (1) has a global AC solution on $[0, +\infty)$.
$\bullet$ However, in Vidyasagar's book [3, Theorem (25), p.38], it is only assumed that $\|f(t,x_0)\|$ is bounded for all t and $f(t,\cdot)$ is a Lipschitz continuous function for all t then (1) has a global continuous and everywhere differentiable solution on $[0, +\infty)$.
The main issue is whether we need the measureability of $f(\cdot,x)$ for a given x or not? It is clear that in Khalil's book (piecewise continuous infers measureability and boundedness) and Sontag's need for the measureability of $f(\cdot,x)$, however in Vidyasagar's book this is not needed.
Is this an omission on the author's part? Because the proof of this result involves Picard's iterations where it is written $\int_{0}^{t} f(\tau, x_0)d\tau$. I think at least for this integral to exist then $f(\cdot, x_0)$ must be measurable. What do you think?
Also note that: The solutions in Vidyasagar's book are not AC (because a function that is continuous and differentiable everywhere is not necessarily AC). This is a solution with weaker properties than AC. What explanation could this possibly have for not mentioning the measure of $f(\cdot, x)$?
[2] http://www.sontaglab.org/FTPDIR/sontag_mathematical_control_theory_springer98.pdf