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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?

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

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, $$\chi_{E}^{-1}\left(\frac{1}{2}, \frac{3}{2}\right)=E\notin\mathcal{B}_{R}$$

Remark. I am using notation consistent with Folland's Real Analysis: Modern Techniques and Their Applications.

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