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Let $\left(X,\mathcal{A},\mu \right)$ be a measure-space and $f:X\to \mathbb{\left[}0,\mathrm{\infty }\right]$ a unsigned measurable map. Show that $\mu \left(\left\{f\left(x\right)>0\right\}\right)=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mu \left(\left\{f\left(x\right)>t\right\}\right)=0$ For every $t>0$.
Suppose that $\mu \left(\left\{f\left(x\right)>0\right\}\right)=0$. Since $f=0$ a.e we have that ${\int }_{X}f=0$. Thus $\mu \left(\left\{f\left(x\right)>t\right\}\right)\le \frac{1}{t}{\int }_{X}f=0$ and since the measure is positive we have that $\mu \left(\left\{f\left(x\right)>t\right\}\right)=0$.
Conversely suppose that $\mu \left(\left\{f\left(x\right)>t\right\}\right)=0$. Then what can be done for this case? It seems that I could write ${\int }_{X}f={\int }_{X\setminus \left\{f\left(x\right)>t\right\}}f+{\int }_{\left\{f\left(x\right)>t\right\}}f$ and the latter integral would be zero, but I don’t really get anywhere from here?
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grindweg1v
Note that $\left\{f\left(x\right)>0\right\}=\bigcup _{n\ge 1}\left\{f\left(x\right)>1/n\right\}$, it holds
$\mu \left(\left\{f\left(x\right)>0\right\}\right)=\underset{n\to \mathrm{\infty }}{lim}\mu \left(\left\{f\left(x\right)>1/n\right\}\right)=0$
since the sequence of set ${A}_{n}=\left\{f\left(x\right)>1/n\right\}$ is increasing.