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Let $ m $ be the Lebesgue measure on $ \mathbb{R} $ and $ f: \mathbb{R} \to [0,\infty) $ a Lebesgue-integrable function.

Show that there exists a Lebesgue-measurable set $ E \subseteq [0,\infty) $ such that $ m(E) \neq m({f^{−1}}[E]) $.

I am totally clueless about how to proceed with this. Any help?

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  • $\begingroup$ If $ f $ is not required to be in $ {\mathcal{L}^{1}}(\mathbb{R}) $, then $$ f(x) = \begin{cases} n + x & \text{if $ n \in \mathbb{Z}_{\geq 0} $ and $ x \in [n,n + 1) $}; \\ |n| - x & \text{if $ n \in \mathbb{Z}_{< 0} $ and $ x \in [n,n + 1) $} \end{cases} $$ would be a counterexample to the OP’s claim. $\endgroup$ Commented Apr 23, 2015 at 8:46

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By way of contradiction, assume that an integrable function $ f: \mathbb{R} \to [0,\infty) $ exists such that $$ \mu(E) = \mu(f^{\leftarrow}[E]) $$ for any Lebesgue-measurable subset $ E $ of $ [0,\infty) $. Then $$ \forall n \in \mathbb{N}: \quad \mu({f^{\leftarrow}}[n,n + 1)) = \mu([n,n + 1)) = 1. $$ Hence, \begin{align} \int_{{f^{\leftarrow}}[1,\infty)} f ~ \mathrm{d}{\mu} & = \sum_{n = 1}^{\infty} \int_{{f^{\leftarrow}}[n,n + 1)} f ~ \mathrm{d}{\mu} \\ & \geq \sum_{n = 1}^{\infty} n \cdot \mu({f^{\leftarrow}}[n,n + 1)) \\ & = \sum_{n = 1}^{\infty} n \cdot \mu([n,n + 1)) \\ & = \sum_{n = 1}^{\infty} n \\ & = \infty. \end{align} This contradicts the hypothesis that $ f \in {\mathcal{L}^{1}}(\mathbb{R}) $.

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  • $\begingroup$ can you please explain the inequality step please ? how is it $\geq $ $\endgroup$
    – Learnmore
    Commented Apr 23, 2015 at 13:14
  • $\begingroup$ @learnmore: Hi. Let $ n \in \mathbb{N} $. For each $ x \in {f^{\leftarrow}}[[n,n + 1)] $, we have $ f(x) \in [n,n + 1) $ by the definition of ‘pre-image’. This means that $ f|_{{f^{\leftarrow}}[[n,n + 1)]} \geq n $. $\endgroup$ Commented Apr 23, 2015 at 14:16
  • $\begingroup$ how should i interpret the fact that a real valued function is Lebesgue Integrable $\endgroup$
    – Learnmore
    Commented Apr 23, 2015 at 15:13
  • $\begingroup$ @learnmore: Hi learnmore. Are you asking what it means for a Lebesgue-measurable function $ f: \mathbb{R} \to \mathbb{R} $ to be Lebesgue-integrable? It simply means that $$ \int_{\mathbb{R}} |f| ~ \mathrm{d}{\mu} < \infty. $$ The integral is taken to be the limit of a sequence of integrals of simple functions that converge to $ |f| $ pointwise from below. If $ f \geq 0 $ in the first place, then this condition simplifies to $$ \int_{\mathbb{R}} f ~ \mathrm{d}{\mu} < \infty. $$ $\endgroup$ Commented Apr 23, 2015 at 16:45
  • $\begingroup$ thanks that was what I am asking $\endgroup$
    – Learnmore
    Commented Apr 23, 2015 at 16:50

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