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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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    I voted to re-open because the duplicate thread contains no answer to the specific question.2012-09-11

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