Absolutely continuous and almost everywhere solution of a controlled dynamical system This is a doubt i had while reading the first chapter form the book Mathematical Control Theory, Jerzy Zabczyk, in the first chapter, page $11$ the author talks about solution for a class of differential equations: 
Consider the system of differential equations $$\frac{dq}{dt}=A(t)q(t)+a(t),~~q(t_0)=q_0 \in \mathbb{R}^n$$
on a fixed interval $[0,T]$; $t_0 \in [0,T]$ where $A \in M(n,n)$ i.e space of all $n \times n$ matrices over $\mathbb{R}$, $A(t)=[a_{ij}, i=1,\ldots,n,j=1,\ldots,m]$, $a(t) \in \mathbb{R}^n$ defined as $a(t)=(a_i(t),i=1,\ldots,n)$, whete $t \in [0,T]$. Now he states the theorem 
Theorem : Assume the elements of the function $A(\cdot)$ are locally integrable. Then there exists exactly one function $S(t)~,~t\in [0,T]$ with values in $M(n,n)$ and with absolutely continuous elements such that $$\begin{align}\frac{d}{dt}S(t)&=A(t)S(t)~~,~~\text{for almost all}~t\in [0,T] \\S(0)&=I \end{align}$$ In addition, a matrix $S(t)$ is invertible for an arbitrary $t \in [0,T]$, and the unique solution of the equation $(1)$ is of the form $$q(t)=S(t)S^{-1}(t_0)q_0+\int_{t_0}^tS(t)S^{-1}(s)a(s)ds,~t\in[0,T]$$ The proof is not that hard to follow, he uses Banach's Fixed point theorem for the firt part , and the second part need some work but doable. But what confuses me are the parts Almost everywhere and absolutely continuous in the statement of the theorem, what does these terms mean actually in this context, i'm not able to follow these two definitions. Will be thankful if someone provides a easy definition for these. Thank you !
 A: The matrix $A$ is a generalized function, its values can be changed on any null set without changing it, that is, without leaving the equivalence class of ordinary functions that make up $A$. Any claims involving values of $A$ can only be "a.e." 
"absolutely continuous" is the strongest regularity you can get in an anti-derivative of a locally integrable function. It replaces the differentiability you would get if $A$ were continuous.
A: I won't go to much detail, for a detailed exposition a course in measure theory would be helpful i feel. Control theory really hinges on the notion of Caratheodory solution of differential equations as opposed to the standard differential equations with Lipschitz right-hand sides, as i said earlier complete understanding of this concept requires measure theory, for our purposes, however, it's enough to know a few things from basic analysis.
In basics of Lebesgue integration theory it is enough to think of a Lebesgue measurable real-valued function on an interval as a piecewise constant function or the pointwise limit of a piecewise constant function. For instance, $$[0,1]∋t↦\text{sgn}(\cos(2πnt))$$ is a Lebesgue measurable discontinuous function for each $n>0$, the Dirichlet comb (i.e the indicator of rationals) $$[0,1]∋t↦\chi_{\mathbb{Q}}(t)=\mathbb{1}_{\mathbb{Q}}(t) := \left\{
\begin{aligned}1  ~~~~~  &t \in\mathbb{Q}\\0~~~~~&\text{else}\end{aligned}
\right. $$ is a Lebesgue measurable function that is "almost everywhere" zero (except a set of measure zero), so it is equivalent to the zero function. If you can inject  $\chi_{\mathbb{Q}}(t)$ as a control input of a system i.e $\mathbb{R} \ni t\mapsto u(t):=\chi_{\mathbb{Q}}(t)$, its response will be identical to its response corresponding to the zero control. Moreover, the set of Lebesgue measurable functions on an interval constitutes a vector space over $\mathbb{R}$ when equipped with pointwise addition and scalar multiplication operations.
An absolutely continuous function is very close to a continuously differentiable function, so it's enough to substitute "continuously differentiable" in place of "absolutely continuous". For instance, the triangle function $$[0,1]∋t↦T(t) := \left\{
\begin{aligned}t  ~~~~~  &\text{if}~~t \in [0,1/2]\\1-t~~~~~&\text{if}~~t \in~]1/2,1]\end{aligned}
\right. $$ is absolutely continuous but not continuously differentiable. The defining characteristic of an absolutely continuous function is that they're expressible as integrals of other, more rough functions. You may view the process of integration be it Lebesgue or Riemann, as a smoothing operation: typically, rough functions become less rough upon integration. Absolutely continuous functions are very close to functions of bounded variation, if you can write down a given function as an integral as in $$f(t)=a+\int_b^t g(s)\,\mathrm{d}s$$ for some other function $g$, then $f$ is absolutely continuous. For instance, for the triangle function, $a=0$, $c=0$ and $$[0,1]∋t↦g(t) := \left\{
\begin{aligned}t  ~~~~~  &\text{if}~~1 \in [0,1/2[\\-1~~~~~&\text{if}~~t \in~[1/2,1]\end{aligned}
\right. $$
