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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?

4 Answers 4

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Hint:

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

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    @brdcastguy: Exactly.2012-11-08
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Hint:

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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    It is indeed...2012-11-08
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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.2012-11-09