# Proving the existence of a certain Lebesgue-measurable set.

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?

• 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. Commented Apr 23, 2015 at 8:46

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 $f \in {\mathcal{L}^{1}}(\mathbb{R})$.
• can you please explain the inequality step please ? how is it $\geq$ Commented Apr 23, 2015 at 13:14
• @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 $f|_{{f^{\leftarrow}}[[n,n + 1)]} \geq n$. Commented Apr 23, 2015 at 14:16
• @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 $$\int_{\mathbb{R}} f ~ \mathrm{d}{\mu} < \infty.$$ Commented Apr 23, 2015 at 16:45