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 $\lambda(A)=0$.

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.