Suppose $f:[a,b]\rightarrow [-\infty, \infty]$ is bounded and Riemann integrable, must it be measurable with respect to the Boreal measure on $[a,b]$?
A Riemann integrable function $f$ on a bounded interval $[a, b]$ is measurable with respect to the Borel measure on $[a,b]$?
5
$\begingroup$
measure-theory
-
0The above 2 comments are true in their respective contexts, which are not mentioned here. For a complete treatment, see:http://math.stackexchange.com/questions/291020/does-riemann-integrable-imply-lebesgue-integrable – 2013-07-03
1 Answers
7
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.)
-
1Oh, 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. – 2012-10-31