# Problem of the Week #12 - August 20th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $f$ be a measurable function on a measure space $(X,\Lambda,\mu)$. Show that the pre-image of any Borel set of $\mathbb{R}$ is also in $\Lambda$.

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#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. Here's my solution.

Proof: Let $f$ be measurable and let $B\subset\mathbb{R}$ be a Borel set. Let $\mathcal{F}=\{E\subset\mathbb{R}:f^{-1}(E)\in\Lambda\}$ be a collection of sets. Assuming that $E\in\mathcal{F}$, then
$f^{-1}(E^c)=(f^{-1}(E))^c\text{ is measurable}.$

Therefore, $E^c\in\mathcal{F}$. Now, suppose $\{E_i\}\in\mathcal{F}$ is a sequence of sets. Then

$f^{-1}(\bigcup E_i)=\bigcup f^{-1}(E_i)\text{ is measurable}.$

Therefore, $\bigcup E_i\in\mathcal{F}$. Thus, $\mathcal{F}$ is a $\sigma$-algebra.

Now, note that for any $a,b\in\mathbb{R}$ with $a<b$, the sets $\{x:f(x)>a\}$ and $\{x:f(x)<b\}$ are both measurable. Thus, it follows that $(a,\infty),(-\infty,b)\in\mathcal{F}$. Thus, $(a,b)=(a,\infty)\cap(-\infty,b)\in\mathcal{F}$. Therefore, the $\sigma$-algebra $\mathcal{F}$ contains all open sets, which implies that all Borel sets $B$ are in $\mathcal{F}$. Thus, $f^{-1}(B)$ is measurable.

Q.E.D.

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