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?
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}) $.