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

lebesgue-integrallebesgue-measuremeasure-theoryqueueing-theory

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 $$\lambda(A)=0$$.

#### Best Answer

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.