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Possible Duplicate:
Are measurable functions closed under addition and multiplication, but not composition?

Let $\varphi: \mathbb{R}\to \mathbb{R}$ and $f$ is a real-valued measurable function. If $\varphi$ is continuous then I can show that $\varphi \circ f$ is measurable. Is the conclusion still true if we only know the $\varphi$ is Lebesgue measurable?

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    Do you mean Borel or Lebesgue measurable?2012-03-31
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    What is your definition of "measurable", and how did you prove the case of $\varphi$ continuous?2012-03-31
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    If you mean Lebesgue measurable, the answer is "no"; see example 8.38 on page 109 of Gelbaum and Olmstead's [Counterexamples in Analysis](http://books.google.com/books/about/Counterexamples_in_Analysis.html?id=cDAMh5n4lkkC). The [wiki page](http://en.wikipedia.org/wiki/Measurable_function) is also edifying.2012-03-31
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    I mean Lebesgue measurable.2012-03-31
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    You would need that the pre-image of every Lebesgue-measurable set under $f$ is still Lebesgue measurable. You only know that the pre-image of a Borel measurable set under $f$ is Lebesgue measurable. The gap is significant, as the references in David Mitra's comment explain.2012-03-31
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    I do not think this is a good duplicate and vote against closing (in particular the suggested duplicate contains no answer to the question asked explicitly here: *is the composition of two Lebesgue measurable functions Lebesgue measurable?*).2012-09-10
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    I voted to re-open because the duplicate thread contains no answer to the specific question.2012-09-11

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