1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Berrick Caleb Fillmore Apr 23 '15 at 8:46
2
$\begingroup$

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}) $.

$\endgroup$
  • $\begingroup$ can you please explain the inequality step please ? how is it $\geq $ $\endgroup$ – Learnmore Apr 23 '15 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 $. $\endgroup$ – Berrick Caleb Fillmore Apr 23 '15 at 14:16
  • $\begingroup$ how should i interpret the fact that a real valued function is Lebesgue Integrable $\endgroup$ – Learnmore Apr 23 '15 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 $$ \int_{\mathbb{R}} f ~ \mathrm{d}{\mu} < \infty. $$ $\endgroup$ – Berrick Caleb Fillmore Apr 23 '15 at 16:45
  • $\begingroup$ thanks that was what I am asking $\endgroup$ – Learnmore Apr 23 '15 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.