$f_n$ 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 $$\lim_{n\to\infty} \int_D f_n(x)\, \mbox{d}x = \int_D f(x)\, \mbox{d}x$$

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    $\begingroup$ 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

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 $\int_D f=0$.


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