In many textbooks, it is often written that a Riemann integrable (bounded) function on a compact interval [a,b] is also Lebesgue-integrable, and the two integrals are equal. My question is that, is measurability of the function (w.r.t. the Borel sigma field) implicitly assumed in these statements, or does it follow automatically from Riemann integrability of the function. I don't think so. Here, I am trying to give a counterexample, and I wish that its truth is confirmed. Let C be the Cantor set. We know that C being uncountable, contains a non-Borel subset, say S. Define f on [0,1] as: f(x)= 1 (if x is in S) and 0 (if x is in [0,1] - S). In other words, f is just the indicator function of S. Clearly, f is not measurable. Now, f being constant on the complement of C, which is open, f is continuous on the complement of C. Since the Lebesgue measure of C is 0, f is continuous Lebesgue-almost-everywhere, hence must be Riemann integrable! Is there any flaw in this counterexample? Any help will be greatly appreciated!

  • 2
    $\begingroup$ Your function is Lebesgue measurable (and Lebesgue integrable). As you show, a Riemann integrable function need not be Borel measurable. This is also addressed here. $\endgroup$ – David Mitra Oct 24 '14 at 12:10
  • $\begingroup$ Thanks a lot, David. Actually, in the general integration theory I learnt, integrals were defined for functions measurable w.r.t the domain sigma algebra only. Now, your writing gives me a feeling that integrals are actually defined in the trivial manner for functions, the inverse image of borel sets under which, belong to the completion of the domain sigma-field. Am I correct? If not, can you please mention the most general requirement on a function to define integrals in abstract measure spaces? $\endgroup$ – Somabha Mukherjee Oct 24 '14 at 12:32
  • $\begingroup$ Yes, every measure $\mu$ on a measure space $(X, \mathcal{M},\mu)$ has a unique completion, that is there is a smallest $\sigma$-algebra $\mathcal{M}' \supset \mathcal{M}$ and a measure $\mu'$ on $\mathcal{M}'$ which is complete, i.e. for every set $N \in \mathcal{M}'$ with $\mu'(N) = 0$ and $N' \subset N$, we have $N' \in \mathcal{M}'$. On can now define the usual integral w.r.t. $\mu '$ and thus integrate any function that you mention. Some people directly define the integral in such a way that the integrable functions are those measurable w.r.t $\mathcal{M}'$ $\endgroup$ – PhoemueX Oct 24 '14 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.