Show $\gamma(t)\leq 0$ for almost all $t$ with $\max_{u\leq t} \int_u^t \gamma \,\mathrm d\lambda = 0$

Question

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 $\lambda(A)=0$.

Answer 1

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.

Answer 2

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 $$ \lambda(A) \leq \lambda((B \cap C)^c) = \lambda(B^c \cup C^c) \leq \lambda(B^c) + \lambda(C^c) = 0 $$