# A Riemann integrable function $f$ on a bounded interval $[a, b]$ is measurable with respect to the Borel measure on $[a,b]$?

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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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. – Beni Bogosel Oct 31 '12 at 15:05
The above comment is false. $f(x) = \frac{\sin(x)}{x}$ is Riemann integrable but not Lebesgue integrable. – madprob Nov 1 '12 at 6:19
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/… – Anonymous Coward Jul 3 '13 at 15:09

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.)
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. – kahen Oct 31 '12 at 15:48
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. – Lukas Geyer Oct 31 '12 at 15:50