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I need to show that the following real valued function on $\mathbb{R}$ is nonmeasurable: $$ f(x) = \begin{cases} x &\text{$x \in E$} \\ -x &\text{$x \in [0,1] \setminus E$} \end{cases} $$

where $E$ is a non-measurable subset of $[0,1]$. How could one do this?

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If the function was measurable then $f^{-1}([0,1])$ would be a measurable set.

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Is $f^{-1}([0,1])$ nonmeasurable because it is equal to $E$? Sorry, I am a little lost. – brdcastguy Nov 8 '12 at 22:35
@brdcastguy: Exactly. – Asaf Karagila Nov 8 '12 at 22:35


Suppose $0 \in E$, then what is the set $\{x | f(x) \geq 0 \}$?

(If $0 \notin E$, then use $f(x) \leq 0$ instead.)

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Would $f^{-1}([0,\infty)) = E$, which is nonmeasurable, so $f$ is nonmeasurable? Is that the conclusion I need to make? – brdcastguy Nov 8 '12 at 22:31
It is indeed... – copper.hat Nov 8 '12 at 23:44

The inverse image of a measurable set under a measurable function must be measurable. Were $f$ to be measurable, $g(x) = f(x) - x$ is also measurable We know that $\{0\}$ is a measurable subset of the line. However, $g^{-1}(\{0\}) = E$ is not measurable. Hence $g$ is not measurable. But if $g$ is not measurable, then $f$ is not.

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How about proving that if $f$ is any measurable function, then so is $$ x\mapsto\frac{f(x)}{x}.\tag{1} $$ For your function $f$, the inverse image of the set $\{1\}$ under the function $(1)$ is $E$, which would therefore be measurable.

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works too. There's a panoply of ways to do this. – ncmathsadist Nov 9 '12 at 0:31

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