Derivative of $\Gamma(t):=\max_{u\leq t} \int_u^t \gamma \,\mathrm d\lambda$

Given a locally integrable function $\gamma:\mathbb R_{\geq0}\rightarrow\mathbb R$ we define the continuuous, non-negative function $\Gamma(t):=\max_{u\leq t}\int_u^t \gamma \,\mathrm d\lambda$. $\Gamma$ can be interpreted as the length of a queue with flow rate $\gamma$.

I want to show, that $\Gamma$ is differentiable almost everywhere with $$ \Gamma'(t) = \begin{cases} \gamma(t), &\text{for $\Gamma(t)>0$,}\\ 0, &\text{otherwise,} \end{cases} $$ for almost all $t\geq 0$.

I could manage to show that $\Gamma'(t) = \gamma(t)$ for almost all $t$ with $\Gamma(t)>0$. I include the details for completeness, although they are not too relevant for the question:

We first show, that on closed intervals $[a,b]$ with $a < b$ on which $\Gamma$ is strictly positive, there exists a common $u\leq a$ such that $\Gamma(\theta) = \int_u^\theta \gamma \,\mathrm d\lambda$ holds for all $\theta\in [a,b]$. Let $u_a\in\arg\max_{u\leq a} \int_u^a \gamma \,\mathrm d\lambda$. We define $$\theta := \max\left\{\theta\in [a, b] \,\middle\vert\, \forall \theta'\in[a,\theta]: \Gamma(\theta') = \int_{u_a}^{\theta'} \gamma \,\mathrm d\lambda \right\}.$$ If $\theta = b$, the claim follows directly. If we assume $\theta < b$, there exists a sequence $(\theta_k)_{k\in\mathbb N}$ with $\theta_k\in(\theta,b]$, $\lim_{k\rightarrow\infty} \theta_k= \theta$ and $\Gamma(\theta_k) \neq \int_{u_a}^{\theta_k} \gamma \,\mathrm d\lambda$. By definition of $\Gamma$ we even have $\Gamma(\theta_k) > \int_{u_a}^{\theta_k} \gamma \,\mathrm d\lambda$. Moreover, this shows that $\Gamma(\theta_k) = \max_{u\in(\theta, \theta_k]} \int_u^{\theta_k} \gamma \,\mathrm d\lambda$, because for $u\leq\theta$ we have $$\int_u^{\theta_k} \gamma \,\mathrm d\lambda \leq \Gamma(\theta) + \int_\theta^{\theta_k} \gamma\,\mathrm d\lambda = \int_{u_a}^{\theta_k} \gamma \,\mathrm d\lambda < \Gamma(\theta_k).$$ Therefore, for $k \rightarrow \infty$ the value $\Gamma(\theta_k)$ converges to $0$. Because $\Gamma$ is continuous we derive the contradiction $\Gamma(\theta) = 0$.

What troubles me is to get the required result for almost all $t$ with $\Gamma(t) = 0$: The set $A:=\{t\mid \Gamma(t) = 0\}$ is closed by the continuity of $\Gamma$. On open intervals $(a,b)\subseteq A$, the derivative is clearly $\Gamma'(t)=0$, but I lack this statement for the boundary $\partial A$. I am unsure whether the boundary is relevant in my case, but there are closed sets with positive measure on the boundary (e.g. a Smith-Volterra-Cantor-Set).

Edit: I had a typo in the question: I previously said, $\gamma$ would be non-negative. In that case, Henry Bosch has given a very straight-forward answer in https://math.stackexchange.com/a/4311863/652841. Also for $\gamma\geq 0$ we have $\Gamma(t) = \int_0^t \gamma \,\mathrm d \lambda$.

This answer assumes that $\gamma \geq 0$. This was originally assumed by OP, but now this has changed.

The set $A := \{t|\Gamma(t) = 0\}$ is an interval. Indeed, if $\Gamma(t_0) = 0$, then $$0 = \Gamma(t_0) = \max_{0 \leq u \leq t_0} \int_u^{t_0} \gamma d\lambda \geq \int_0^{t_0} \gamma d\lambda $$ which forces $\gamma = 0$ a.e. on $[0,t_0]$. If $0 \leq t \leq t_0$, this forces $\Gamma(t) = 0$. Thus $A$ has the property such that if $t_0 \in A$ and $0 \leq t \leq t_0$, then $t \in A$. This implies that $A$ is an interval, so its boundary has measure 0.