Given a Sturm-Liouville type operator which acts on functions on the interval $(a,b)$
$$T:= \frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x) \frac{d}{dx}\right]+q(x)\right)$$
where $w$, $p^{-1}$ and $q \in L_{\text{loc}}(a,b)$ and are real valued, we define its maximal domain as
$$ D_{\text{max}}:=\big \{f \in L^2_w(a,b) : f,pf'\in \text{AC}_{\text{loc}}(a,b), \quad Tf\in L^2_w(a,b) \big \} $$
Where $L^2_w(a,b)$ is the space of square integrable functions on $(a,b)$ with weight $w$.
I am trying to understand the motivation for defining the maximal domain as above.
I understand the requirement that $f$ and $Tf\in L^2_w(a,b)$, since we want $T$ to be an endomorphism (so we can apply the Hilbert space theory of $L^2_w(a,b)$).
However I am not totally certain why we ask for absolute continuity of $f$ and $pf'$.
I surmise that it's because it is the weakest assumption needed to make the derivatives $f'$ and $(pf')'$ Lebesgue integrable (e.g. see Lebesgues fundamental theorem of calculus), which is needed in order to perform integration by parts as to investigate the symmetry of operator $T$. i.e.
\begin{align} \langle Tf,g \rangle &= \int_a^b \left(-\frac{d}{dx}\left[p(x) \frac{df}{dx}\right]+q(x)f\right)\cdot\overline{g} \quad dx=\text{boundary terms } + \langle f,Tg \rangle \end{align}
Does this seem reasonable? I am just concerned I am over looking something, and the texts I am following (Zettls Sturm-Liouville Theory and Naimarks Linear Differential Operators) don't seem to mention why the AC condition is required.
EDIT:
An equivalent definition we could give of $D_{\text{max}}$ is that its the domain of the maximal operator, however I'd be happier with a more apriori definition though, as the maximal domain seems to be introduced in the referenced texts before the introduction of the minimal or maximal operators.
A typical minimal domain for $T$ would consist of $C^{\infty}_{c}(a,b)$, which consists of infinitely differentiable functions that are compactly supported in $(a,b)$. The minimal operator $T_m$ with this domain then gives rise to the maximal operator $T_{M}=T_m^*$, which is the adjoint of the minimal operator. This works because $T_{m}$ is symmetric on its domain. So the maximal operator $T_M$ is an extension of $T_m$, but it is maximal because the graph of $T_M$ is the orthogonal complement of the graph of $T_m$ in $L^2(a,b)\times L^2(a,b)$. Defining it in this way basically ensures that the maximal operator is the largest closed extension of $T_m$ in $L^2(a,b)\times L^2(a,b)$. This keeps the discussion in a Hilbert space. You could extend beyond the Hilbert space into distributions, but this keeps the discussion in a Hilbert space. The maximal domain has no endpoint restrictions, unlike the minimal domain where functions vanish near the endpoints.