Show $\gamma(t)\leq 0$ for almost all $t$ with $\max_{u\leq t} \int_u^t \gamma \,\mathrm d\lambda = 0$ Given a locally integrable function $\gamma: \mathbb R_{\geq0}\rightarrow \mathbb R$, we define the absolutely continuous function $\Gamma(t) := \max_{u\leq t} \int_u^t \gamma \,\mathrm d\lambda$.
I want to show, that $\gamma(t)\leq 0$ holds for almost all $t$ with $\Gamma(t)=0$.
In other words, I want to show that the set $$
A :=  \left\{ t\in\mathbb R_{\geq 0} \,\middle\vert\, \Gamma(t) = 0 ~\text{and}~ \gamma(t) > 0 \right\}
$$
is a Lebesgue-null set, i.e. $\lambda(A)= 0$.
All my attempts have failed. Nevertheless, I was able to show, that if $\Gamma$ vanishes on a (proper) interval $[a,b]$ with $a < b$, then $\lambda(A\cap [a,b]) = 0$.
This however does not lead to a proof of the more general claim $\lambda(A)=0$.
 A: Let $B$ be the following measurable set.
\begin{equation*}
B = \left\{t \in (0,\infty) \, \mid \, \lim_{u \to t^{-}} \frac{1}{t - u} \int_{u}^{t} \gamma(s) \, ds = \gamma(t)\right\}.
\end{equation*}
By the differentiation theorem, $\mathbb{R}_{\geq 0} \setminus B$ is a Lebesgue null set.  I leave it to you to show that $\Gamma > 0$ holds in the measurable set $B \cap \{\gamma > 0\}$.  Hence $\Gamma > 0$ holds Lebesgue almost everywhere in $\{\gamma > 0\}$, or $\{\Gamma = 0, \, \, \gamma > 0\}$ is Lebesgue null.
A: We first define two measurable sets:
$$B:=\left\{t \in \mathbb R_{\geq0} \,\middle\vert\, x\mapsto\int_0^x \gamma\,\mathrm d\lambda \text{ is differentiable in $t$ with derivative $\gamma(t)$}  \right\}$$
$$ C:= \left\{t \in \mathbb R_{\geq0} \,\middle\vert\, \text{$\Gamma$ is differentiable in $t$ with $\Gamma'(t)=\begin{cases}
\gamma(t), &\text{for $\Gamma(t) > 0$,}\\
0, &\text{otherwise.}
 \end{cases}$} \right\} $$
By Lebesgue's differentiation theorem, we know that $\lambda(B^c) = 0$.
By the proof in this answer, we also have $\lambda(C^c) = 0$.
Now let $t\in B\cap C$ with $\Gamma(t) = 0$.
Then we have $$
\left. \frac{\,\mathrm d}{\,\mathrm d x} \int_t^x \gamma\,\mathrm d\lambda \,\right\vert_{x=t} = 
\left. \frac{\,\mathrm d}{\,\mathrm d x} \int_0^x \gamma\,\mathrm d\lambda - \int_0^t \gamma\,\mathrm d\lambda \,\right\vert_{x=t} = \gamma(t)
$$
and thus it holds that
$$
0 = \Gamma'(t) = \lim_{x\,\searrow\, t} \frac{\Gamma(x) - \Gamma(t)}{x-t} =
\lim_{x\,\searrow\, t} \frac{\Gamma(x) - \int_t^t \gamma\,\mathrm d\lambda}{x-t}
\geq \lim_{x\,\searrow\, t} \frac{\int_t^x \gamma \,\mathrm d\lambda - \int_t^t \gamma \,\mathrm d\lambda}{x-t}
= \left. \frac{\,\mathrm d}{\,\mathrm d x} \int_t^x \gamma\,\mathrm d\lambda \,\right\vert_{x=t} = \gamma(t).
$$
This means $A\cap B \cap C = \emptyset$ and thus $A\subseteq (B\cap C)^c$.
Moreover, we have $$
\lambda(A) 
\leq \lambda((B \cap C)^c) = \lambda(B^c \cup C^c)
\leq \lambda(B^c) + \lambda(C^c) = 0
$$
