F non-measurable but |f| and f^2 are

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SUMMARY

The discussion focuses on identifying a function within a measurable space that is non-measurable while ensuring that both its absolute value, |f|, and its square, f², are measurable. The key conclusion is that for any subset S of the real numbers, the preimage under |f| and f² results in either the entire space X or the empty set, depending on whether S includes the value 1. This establishes that |f| and f² maintain measurability despite f being non-measurable.

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Homework Statement



Find a function in a measurable space that is non-measurable, but |f| and f2 are measurable.

Homework Equations



None.

The Attempt at a Solution



I am trying to understand the following answer to the problem:

MMx7fFs.png

(source: http://math.stackexchange.com/a/1233792/413398)

I do not understand why for all subsets S \subseteq \mathbb{R}, we have $$(|f|)^{-1}(S)=(f^2)^{-1}(S)=X \text{ or } \varnothing$$
 
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Hi,

##f^2## or ##abs(f)## both send ##a## and ##b## to ##1##. So ##1## has ##a## and ##b## as antecedents, and any other real has no antecedent. So if ##S## contains ##1## it has ##X=\{a,b\}## as reciprocal image, and the empty set if it doesn't contain it.
 
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