# I canr do this, need help Given a locally integrable function &#x03B3;<!-- γ --> :

I canr do this, need help
Given a locally integrable function $\gamma :{\mathbb{R}}_{\ge 0}\to \mathbb{R}$, we define the absolutely continuous function $\mathrm{\Gamma }\left(t\right):=\underset{u\le t}{max}{\int }_{u}^{t}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda$.
I want to show, that $\gamma \left(t\right)\le 0$ holds for almost all $t$ with $\mathrm{\Gamma }\left(t\right)=0$. In other words, I want to show that the set

is a Lebesgue-null set, i.e. $\lambda \left(A\right)=0$.
All my attempts have failed. Nevertheless, I was able to show, that if $\mathrm{\Gamma }$ vanishes on a (proper) interval $\left[a,b\right]$ with $a, then $\lambda \left(A\cap \left[a,b\right]\right)=0$. This however does not lead to a proof of the more general claim $\lambda \left(A\right)=0$.
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Marlee Norman
Let B be the following measurable set.
$B=\left\{t\in \left(0,\mathrm{\infty }\right)\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}\underset{u\to {t}^{-}}{lim}\frac{1}{t-u}{\int }_{u}^{t}\gamma \left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds=\gamma \left(t\right)\right\}.$
By the differentiation theorem, ${\mathbb{R}}_{\ge 0}\setminus B$ is a Lebesgue null set. I leave it to you to show that $\mathrm{\Gamma }>0$ holds in the measurable set $B\cap \left\{\gamma >0\right\}$. Hence $\mathrm{\Gamma }>0$ holds Lebesgue almost everywhere in $\left\{\gamma >0\right\}$, or $\left\{\mathrm{\Gamma }=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\gamma >0\right\}$ is Lebesgue null.
###### Not exactly what you’re looking for?
minwaardekn
We first define two measurable sets:

By Lebesgue's differentiation theorem, we know that $\lambda \left({B}^{c}\right)=0$.
Now let $t\in B\cap C$ with $\mathrm{\Gamma }\left(t\right)=0$. Then we have
${\frac{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}}{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}{\int }_{t}^{x}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda \phantom{\rule{thinmathspace}{0ex}}|}_{x=t}={\frac{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}}{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}{\int }_{0}^{x}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda -{\int }_{0}^{t}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda \phantom{\rule{thinmathspace}{0ex}}|}_{x=t}=\gamma \left(t\right)$
and thus it holds that
$0={\mathrm{\Gamma }}^{\prime }\left(t\right)=\underset{x\phantom{\rule{thinmathspace}{0ex}}↘\phantom{\rule{thinmathspace}{0ex}}t}{lim}\frac{\mathrm{\Gamma }\left(x\right)-\mathrm{\Gamma }\left(t\right)}{x-t}=\underset{x\phantom{\rule{thinmathspace}{0ex}}↘\phantom{\rule{thinmathspace}{0ex}}t}{lim}\frac{\mathrm{\Gamma }\left(x\right)-{\int }_{t}^{t}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda }{x-t}\ge \underset{x\phantom{\rule{thinmathspace}{0ex}}↘\phantom{\rule{thinmathspace}{0ex}}t}{lim}\frac{{\int }_{t}^{x}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda -{\int }_{t}^{t}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda }{x-t}={\frac{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}}{\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}{\int }_{t}^{x}\gamma \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\lambda \phantom{\rule{thinmathspace}{0ex}}|}_{x=t}=\gamma \left(t\right).$
This means $A\cap B\cap C=\mathrm{\varnothing }$ and thus $A\subseteq \left(B\cap C{\right)}^{c}$. Moreover, we have
$\lambda \left(A\right)\le \lambda \left(\left(B\cap C{\right)}^{c}\right)=\lambda \left({B}^{c}\cup {C}^{c}\right)\le \lambda \left({B}^{c}\right)+\lambda \left({C}^{c}\right)=0$