# Lebesgue and Borel Measurable

If a real-valued function on $R$ is measurable with respect to the $\sigma$-algebra of Lebesgue measurable sets, is it necessarily measurable with respect to the Borel measurable space ($R$, $B(R)$)?

I don't think so. Is there a counter example that would be sufficient to show that there isn't?

• Which sigma-algebra(s) on the source set? – Did Mar 14 '13 at 16:46
• math.stackexchange.com/questions/20421/… – Willie Wong Mar 14 '13 at 17:01
• A simple counterexample is the characteristic function of any non-Borel-measurable, Lebesgue-measurable set. – Yoni Rozenshein Mar 14 '13 at 17:16
• the $\sigma$-algebra of Lebesgue measurable sets are just the completion of Borel measurable sets. – Qijun Tan Mar 14 '13 at 17:21
• See problem 35 in Chapter 1 of Real Analysis by E.M. Stein. There is an explicit construction. – Qijun Tan Mar 14 '13 at 17:42

Notation: Let $\mathcal{L}$ and $\mathcal{B}_{\mathbb{R}}$ be the set of Lebesgue measurable sets and Borel measurable sets, respectively.
The answer to your question is 'no'. Take a Lebesgue measurable set $E$ that is not Borel measurable (such sets exist: see here). Consider the characteristic function $\chi_{E}: \mathbb{R}\to\mathbb{R}$ defined by $$\chi_{E} = \begin{cases} 1 & \textrm{ if } x\in E \\ 0 & \textrm{ if } x\notin E \end{cases}$$ Then $\chi_{E}$ is $(\mathcal{L}, \mathcal{B}_{\mathbb{R}})$-measurable (i.e. Lebesgue measurable) because it is characteristic function of a measurable set.
But $\chi_{E}$ is not $(\mathcal{B}_{\mathbb{R}}, \mathcal{B}_{\mathbb{R}})$-measurable (i.e. Borel measurable). For example, $$\chi_{E}^{-1}\left(\frac{1}{2}, \frac{3}{2}\right)=E\notin\mathcal{B}_{R}$$