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The problem is not properly stated in the original, here is the correct version: If $f_n\leq g_n \leq h_n$, $f_n \to f$, $h_n \to h$ pointwise and $\int f_n\to\int f$,$\int h_n \to \int h$ then $\int g_n \to \int g$

Really do not know to how to tackle this problem. it is easy to show that $g \in L^1$. But not sure how to show convergence of integrals. The only thing I see is that $|g_n|\leq \max\{-f_n,h_n\}$ but that is not very helpful. I tried applying generalized dominated convergence theorem but could not succeed. Not sure what else to try. I was wondering if perhaps these conditions somehow imply that all of them converge in $L^1$ Any hints or solutions would be appreciated.

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  • $\begingroup$ What does $\int g_n \to g$ mean? Where is this taken from? $\endgroup$
    – copper.hat
    May 8 '20 at 2:51
  • $\begingroup$ @copper.hat I assume that is an error in the original. The title has it correct. Let me fix this. $\endgroup$
    – Sorfosh
    May 8 '20 at 2:52
  • $\begingroup$ What book is it from? $\endgroup$
    – copper.hat
    May 8 '20 at 2:52
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    $\begingroup$ @copper.hat I have seen it on two past quals form different schools, so no book reference unfortunately. $\endgroup$
    – Sorfosh
    May 8 '20 at 2:53
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$0 \le g_n - f_n \le h_n -f_n$, so $g_n-f_n$ is dominated by the integrable $h_n-f_n$. Hence $g-f$ is integrable and $\int(g_n-f_n) \to \int (g-f)$. Adding $f_n, f$ to each side gives the desired result.

Alternative:

To original intent here was to avoid the generalised DCT but it just ended up being a marginal variation of said theorem.

We have $\int \varliminf (g_n -f_n) \le \int \varliminf (h_n -f_n) \le \varliminf \int (h_n -f_n) = \int (h-f)$. Since $\varliminf (g_n-f_n) = g-f$ ae. we see that $g \in L^1$.

Note that $\int \varliminf (g_n -f_n) \le \varliminf \int (g_n-f_n) = \varliminf (\int g_n - \int f)$ which gives $\int g \le \varliminf \int g_n$. Similarly, $\int \varliminf (h_n -g_n) \le \varliminf \int (h_n-g_n) = \varliminf (\int h - \int g_n)$ which gives $\varlimsup \int g_n \le \int g$.

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  • $\begingroup$ wow, cant believe I did not think of that. Thank you very much. I was wracking my brain on this problem for a while, over complicating it.... Could you tell me what made you think to do that, or did you just instinctively do it and it worked? $\endgroup$
    – Sorfosh
    May 8 '20 at 3:00
  • $\begingroup$ I just looked for a way to apply the DCT. Since I didn't have sign information, I looked to see if I could make it non negative. $\endgroup$
    – copper.hat
    May 8 '20 at 3:01
  • $\begingroup$ For the DCT don't we need that the sequence is dominated by a fixed integrable function? $\endgroup$ May 8 '20 at 3:04
  • $\begingroup$ @stochasticboy321 he probably means Generalized DCT in this case as i mention in my question. $\endgroup$
    – Sorfosh
    May 8 '20 at 3:04
  • $\begingroup$ @Sorfosh Oh cool. I've either forgotten this or never read it, so thanks for the name of the theorem! $\endgroup$ May 8 '20 at 3:05

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