cyfwelestoi

2022-05-25

Let $(X,\mathcal{A},\mu )$ be a measure-space and $f:X\to \mathbb{[}0,\mathrm{\infty}]$ a unsigned measurable map. Show that $\mu (\{f(x)>0\})=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mu (\{f(x)>t\})=0$ For every $t>0$.

Suppose that $\mu (\{f(x)>0\})=0$. Since $f=0$ a.e we have that ${\int}_{X}f=0$. Thus $\mu (\{f(x)>t\})\le \frac{1}{t}{\int}_{X}f=0$ and since the measure is positive we have that $\mu (\{f(x)>t\})=0$.

Conversely suppose that $\mu (\{f(x)>t\})=0$. Then what can be done for this case? It seems that I could write ${\int}_{X}f={\int}_{X\setminus \{f(x)>t\}}f+{\int}_{\{f(x)>t\}}f$ and the latter integral would be zero, but I don’t really get anywhere from here?

Suppose that $\mu (\{f(x)>0\})=0$. Since $f=0$ a.e we have that ${\int}_{X}f=0$. Thus $\mu (\{f(x)>t\})\le \frac{1}{t}{\int}_{X}f=0$ and since the measure is positive we have that $\mu (\{f(x)>t\})=0$.

Conversely suppose that $\mu (\{f(x)>t\})=0$. Then what can be done for this case? It seems that I could write ${\int}_{X}f={\int}_{X\setminus \{f(x)>t\}}f+{\int}_{\{f(x)>t\}}f$ and the latter integral would be zero, but I don’t really get anywhere from here?

grindweg1v

Beginner2022-05-26Added 12 answers

Note that $\{f(x)>0\}=\bigcup _{n\ge 1}\{f(x)>1/n\}$, it holds

$\mu (\{f(x)>0\})=\underset{n\to \mathrm{\infty}}{lim}\mu (\{f(x)>1/n\})=0$

since the sequence of set ${A}_{n}=\{f(x)>1/n\}$ is increasing.

$\mu (\{f(x)>0\})=\underset{n\to \mathrm{\infty}}{lim}\mu (\{f(x)>1/n\})=0$

since the sequence of set ${A}_{n}=\{f(x)>1/n\}$ is increasing.

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