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Suppose $E$ subset $R$ ($R $\is real numbers) where $E$ is Lebesgue measurable, and $f:E\to R$ and defined $g: R\to R$ by \begin{equation*} g(x) = \begin{cases} f(x) & x \in E \\ 0 & x \notin E \end{cases} \end{equation*}

Show that $f$ is measurable then so is $g$.

I came to this question in my self study class and was wondering how would you do such a proof. I spoke to my TA and said do not worry about it. We are not going to do such proofs like that. I was still thinking how though. Can someone please give an outline?

Here is some of the things I know from looking at it. The theorem I guess we can use is suppose E subset A subset R where A is measurable then E is measurable and X is measurable.

Would E need boundaries for f(x) and 0 so g(x) can be measurable? Can someone please show this. I would like to know what it would look like as I am a self learning person.

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How about $g=f 1_{E}$. Products of measurable functions are measurable.

Here is another, more explicit, approach. Note that if $E$ is a measurable set, then its characteristic function $1_E$ is measurable. \begin{eqnarray} \{x | g(x) > c \} &=& \{x \in E | g(x) > c \} \cup \{x \in E^c | g(x) > c \} \\ &=& \{x \in E | f(x) > c \} \cup \{x \in E^c | 1_{E}(x) > c \} \\ &=& (f^{-1}(c,\infty) \cap E) \cup (1_E^{-1}(c,\infty) \cap E^c) \end{eqnarray} All of the sets on the right hand side are measurable, hence so is the left hand side. Since this is the case for all $c$, it follows that $g$ is, by definition, measurable.

(Aside: Not that it matters, but $1_E^{-1}(c,\infty) \cap E^c = E^c$ if $c<0$, and $1_E^{-1}(c,\infty) \cap E^c = \emptyset$ if $c\ge 0$.)

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  • $\begingroup$ Would that be the proof though. Sorry I am new to this and still trying to learn this even though I was told not to worry about it. Is this saying that anything greater and equal to 1 or less than 1 can be measurable because the sum and products of this are measurable $\endgroup$ – 9959 Mar 20 '13 at 6:01
  • $\begingroup$ No. If $f_1,f_2$ are measurable functions, then so is $x \mapsto f_1(x) f_2(x)$. $\endgroup$ – copper.hat Mar 20 '13 at 6:06
  • $\begingroup$ @9959: I added an alternative elaboration. $\endgroup$ – copper.hat Mar 20 '13 at 6:15
  • $\begingroup$ I see. Since f1, f2 are measurable so is the product which makes g measurable. Would this work for the sum too? g= f1e + f2e. $\endgroup$ – 9959 Mar 20 '13 at 6:16
  • $\begingroup$ Any continuous function will work, sum, product, $\max$, $\min$,... $\endgroup$ – copper.hat Mar 20 '13 at 6:18
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For any $c\geq 1$, $\{x:g(x)>c\}=\{x\in E:f(x)>c\}$ measurable.

For any $c<1$, $\{x:g(x)>c\}=\{x\notin E:g(x)=1\}\cup\{x\in E:f(x)>c\}$ measurable.

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Let $A \subseteq \mathbb{R}$ be measurable. If $0\in A$, then $g^{-1} A = E^c \cup f^{-1} A$, otherwise $g^{-1} A = f^{-1} A$. Either way, $g^{-1} A$ is measurable.

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