integrand of the middle line of (\[acbd\]) so (\[condmin\]) would also have to be an equality with this additional term. Therefore the applicability of the theorem is somehow limited as in practice (\[condmin\]) may be satisfied only if the function is “nice" (i.e. its superdifferential should really only consist of $p_1$). Still it applies in some cases where other results fail (see Remarks \[remliyo\] and \[rm:controesempio\]). Many issues are not fully resolved yet and we plan to work on them in the future.

The first statement ($v\le V$) follows from Hypothesis \[D4deterministiccomparison\], it remains to prove second one. The function $$\left\lbrace
\begin{array}{l}
[t,T]\to \mathcal{H}\times\mathbb{R}\\
s \mapsto (b(s,x_{t,x}(s),u(s)), L(s,x_{t,x}(s),u(s))
\end{array}
\right.$$ in view of Hypotheses \[hpD2onb\] and \[hpD3onLandh\] is in $L^1(t,T;\mathcal{H}\times\mathbb{R})$ (in fact it is bounded). So the set of the right-Lebesgue points of this function that in addition satisfy (\[condE\]) is of full measure. We choose $r$ to be a point in this set. We will denote $y= x_{t,x}(r)$.

Consider now two functions $\varphi^{r,y}\in test1$ and $g^{r,y}\in
test2$ such that (we will avoid the index $^{r,y}$ in the sequel) $v\leq \varphi +g$ in a neighborhood of $(r,y)$, $v(r,y) -
\varphi(r,y) - g(r,y) =0$,$(\partial_t)(\varphi+g)(r,y))=q(r)$, $D\phi(r,y)=p_1(r)$ and $D g(r,y)=p_2(r)$. Then for $\tau\in(r,T]$ 