0
$\begingroup$

$f_n$ are uniformly-bounded functions that converge pointwise on a dense subset of the compact set $D$ to a continuous function $f$. Is it true that $$\lim_{n\to\infty} \int_D f_n(x)\, \mbox{d}x = \int_D f(x)\, \mbox{d}x$$

$\endgroup$
  • 1
    $\begingroup$ If the dense subset has measure $0$, would you expect that to imply something about the convergence of the integrals? $\endgroup$ – Daniel Fischer Nov 24 '14 at 13:15
  • $\begingroup$ @DanielFischer No... $\endgroup$ – guest Nov 24 '14 at 13:22
  • $\begingroup$ The answer is no because the functions $f_n(x)$ can be zero out of a dense set with measure zero, so that the integral on the left is always zero while on the right it can be different from zero. I meant to say something more specific that I wrote here $\endgroup$ – guest Nov 24 '14 at 13:47
1
$\begingroup$

No. If $D=[1,0]$, $C\subset D$ is a Cantor set of positive measure (like this one), $f\equiv0$ and $f_n=\chi_C$ for all $n$, then the conditions are satisfied (since $D\setminus C$ is dense in $D$) but $\int_Df_n$ is the measure of $C$ for all $n$ whereas $\int_D f=0$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.