# Confusion about Lebesgue Measurable Functions

I'm a bit confused about the definition of a Lebesgue measurable function. On one hand we have the definition specific to Lebesgue measure:

$f$ is Lebesgue measurable if for every open set U in $\mathbb{R}$, $f^{-1}$(U) is Lebesgue-measurable.

On the other hand we have the general definition of a measurable function:

$f:(X,\Sigma)\rightarrow (Y,\Delta)$ is measurable if for every measurable set $A$ (i.e. $A \in \Delta$), $f^{-1}(A) \in \Sigma$.

Consider the second definition above, in the measure space $(\mathbb{R},\Sigma)$ where $\Sigma$ is the set of Lebesgue-measurable sets. Let $m$ denote the Lebesgue measure. So a Lebesgue-measurable function is not necessarily $m$-measurable, since not every Lebesgue-measurable set is open/Borel?

Any inputs are highly appreciated.

• I think you mean $f^{-1}(A)\in\Sigma$ whenever $A\in\Delta$. – Aweygan Mar 3 '17 at 0:49
• Thanks Aweygan - I've changed the question to fix that. – Simon Li Mar 3 '17 at 0:50

## 1 Answer

One problem with the "general definition" is that a continuous function might not be measurable. There is, for example, a continuous injective function $f$ from $\mathbb R$ to $\mathbb R$ that is a homeomorphism from a closed set $S$ of positive measure to the Cantor set $E$ (which has measure $0$). For any $V \subseteq S$, $V = f^{-1}(f(V))$, but $f(V) \subseteq E$ has measure $0$ and is therefore measurable. Take a non-measurable subset $V$ of $S$ (e.g. the intersection of $S$ with a Bernstein set) and conclude that $f$ is not measurable.