Suppose $f:[a,b]\rightarrow [-\infty, \infty]$ is bounded and Riemann integrable, must it be measurable with respect to the Boreal measure on $[a,b]$?
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$\begingroup$ A Riemann integrable function is Lebesgue integrable and hence it is Lebesgue measurable. I'm not sure about the fact that $f$ is Borel measurable. As you can see here, there are functions which are Lebesgue measurable, but not Borel measurable. $\endgroup$– Beni BogoselOct 31, 2012 at 15:05
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$\begingroup$ The above comment is false. $f(x) = \frac{\sin(x)}{x}$ is Riemann integrable but not Lebesgue integrable. $\endgroup$– madprobNov 1, 2012 at 6:19
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$\begingroup$ The above 2 comments are true in their respective contexts, which are not mentioned here. For a complete treatment, see:math.stackexchange.com/questions/291020/… $\endgroup$– Anonymous CowardJul 3, 2013 at 15:09
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1 Answer
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The answer is no. We know that a function is Riemann integrable iff it is bounded and a.e. continuous. So if you take $f$ to be the characteristic function of a non-Borel set contained in the standard $1/3$-Cantor set (these sets exist by axiom of choice and a neat construction), then $f$ is Riemann integrable but not Borel measurable. (It is Lebesgue-measurable, though.)
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3$\begingroup$ The existence of non-Borel subsets of the standard Cantor set doesn't require Choice, if I recall correctly. The cardinality of the Borel algebra is $\mathfrak c$ (see for example this PDF) but there are $2^{\mathfrak c}$ subsets of the standard Cantor set. $\endgroup$– kahenOct 31, 2012 at 15:48
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1$\begingroup$ Oh, maybe that is an alternate proof. The one I know uses the existence of non-Lebesgue measurable sets and the invariance of the Borel $\sigma$-algebra under homeomorphisms. $\endgroup$ Oct 31, 2012 at 15:50