Let $(f_n)$ and $(g_n)$ be a sequence of integrable functions with $|f_n|\leq |g_n|$ for all $n$. Moreover, $g_n\to g$ and $f_n\to f$ as $n$ goes to infinity. Also assume $g$ is integrable and $\int_n g_n \to \int g$. Prove or disprove $\lim_{n\to\infty}\int f_n(x) dx= \int f(x) dx$.

I have tried to use dominated convergence but the problem is gn are not bounded necessarily. I've tried to come up with a counterexample where the limit oscillates... but again to no avail. Please help.

  • $\begingroup$ I think you need $g$ integrable or else you can take $g_n = f_n$, and the result is not necessarily true. $\endgroup$ – mathworker21 Dec 17 '16 at 20:39
  • $\begingroup$ Hi, assuming g integral, how would we prove the result? Thanks. $\endgroup$ – user172377 Dec 17 '16 at 20:43

Assuming $g$ is integrable. Just run through the proof of dominated convergence Theorem. By Fatou,

$\int g - \int f = \int (g-f) = \int \liminf_n (g_n-f_n) \le \liminf_n \int (g_n-f_n) = \liminf_n (\int g_n - \int f_n) $ $= \int g - \limsup_n \int f_n$ (since $\int g_n \to \int g)$. So, $\int f \ge \limsup_n \int f_n$. Similarly,

$\int g + \int f = \int (g+f) = \int \liminf_n (g_n+f_n) \le \liminf_n \int (g_n+f_n) = \liminf_n (\int g_n + \int f_n)$ $= \int g + \liminf_n \int f_n$. Therefore, $\liminf_n \int f_n \ge \int f$.

  • $\begingroup$ hi thanks, how do we go from lim inf to lim sup? $\endgroup$ – user172377 Dec 17 '16 at 20:59
  • $\begingroup$ Hi, it is a well known fact that for a sequence $(a_n)_n$ that $\liminf_n (-a_n) = -\limsup_n a_n$ $\endgroup$ – mathworker21 Dec 17 '16 at 20:59
  • $\begingroup$ @Socchi See properties of $\liminf$ and $\limsup$. Great answer! $\endgroup$ – Fimpellizieri Dec 17 '16 at 21:09
  • $\begingroup$ hi, thanks you did not change the sign in the proof though? $\endgroup$ – user172377 Dec 17 '16 at 22:28
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    $\begingroup$ otherwise what a great proof, thanks! $\endgroup$ – user172377 Dec 17 '16 at 23:10

EDIT: This was written when the body of the question did not include the assumption $\int g_n \to \int g$.


$$f_n(x)=g_n(x)=\left\{ \begin{array}{lr} n,&x\in[0,1/n)\\ 0,&\text{otherwise} \end{array}\right.$$

Then $f_n=g_n\to0$, and $\int f_n =1$ for all $n$ but $\int f = 0$.

  • $\begingroup$ Notice $g$ is integrable. $\endgroup$ – Fimpellizieri Dec 17 '16 at 20:52
  • $\begingroup$ In the title of the question, it says $\int g_n \to \int g$. $\endgroup$ – mathworker21 Dec 17 '16 at 20:54

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