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I am given that $(\Omega, \Sigma, \mu)$ is a measure space. Also I am given that $f$ is a nonnegattive $\Sigma$-measurable function on $\Omega$.

Now let a function $g:\mathbb{R} \rightarrow[0,\infty]$ and $g(t)=\mu(\{f>t\})$ if $t\geq0$ and $g(t)=0$ if $t<0$.

I need show that $g$ is Lebesgue measurable?

Here are my thoughts...

I tried to show that the set $\{f>t\}$ is Lebesgue measurable. So for any set $A$,

$\mu^{*}(A)$

$=\inf \{ \sum_{k=1}^{\infty} {l(I_{k}):\{ I_{k} \}} \textrm{ is a sequence of open intervals that cover }A \}$

$=\inf \{ \sum_{k=1}^{\infty} {l(I_{k,1} \cup I_{k,2}):\{ I_{k} \}} \textrm{...} \}$

where $I_{k,1}=I_{k}\cap\{f>t\}$ and $I_{k,2}=I_{k}\cap\{f>t\}^c$

and then I can separate the length of the union into the sum of the lengths and hence equal to $\mu^{*}(A \cap \{f>t\})+\mu^{*}(A \cap \{ f>t \}^{c})$

Is my idea correct? Is this the right way to show a function is Lebesgue measurable? Any advice? What about the "$g(t)=0$"?

LanaDR
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1 Answers1

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For convenience let $h:\mathbb R\to[0,\infty]$ be the function prescribed by $t\mapsto\mu(\{f>t\}$.

Note that $g$ is actually the function prescribed by $t\mapsto0$ if $t<0$ and $t\mapsto h(t)$ otherwise, i.e.:$$g(t)=h(t)1_{[0,\infty)}(t)$$

It is not difficult to see that $g$ will be Lebesgue measurable if $h$ is Lebesgue measurable, so let us focus on $h$ from here.

Then evidently $t_1<t_2\implies\{f>t_2\}\subseteq\{f>t_1\}\implies h(t_1)\geq h(t_2)$.

We observed that $h$ is decreasing so that sets like $\{h<x\}$ are intervals or belong to $\{\varnothing,\mathbb R\}$.

That tells us directly that $h$ is Lebesgue measurable.

drhab
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  • Is it obvious to say that ${h<x}$ are intervals? – LanaDR Oct 03 '17 at 09:45
  • If $t\in{h<x}$ then also $t'\in{h<x}$ for any $t'>t$. That makes it possible to prove that ${h<x}$ must be a set of the form $(r,\infty)$, $[r,\infty)$, $\mathbb R$ or $\varnothing$. These sets are Lebesgue measurable so we have found that ${h<x}$ is Lebesgue measurable for (any) $x$. This allows the conclusion that $h$ is a Lebesgue measurable function. – drhab Oct 03 '17 at 09:52
  • Great. Your explanation is very clear and I fully understand now. Thanks! – LanaDR Oct 03 '17 at 10:01
  • You are welcome. – drhab Oct 03 '17 at 10:03