# 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 prove that $\int g_n \to \int g$

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.

• What does $\int g_n \to g$ mean? Where is this taken from? May 8 '20 at 2:51
• @copper.hat I assume that is an error in the original. The title has it correct. Let me fix this. May 8 '20 at 2:52
• What book is it from? May 8 '20 at 2:52
• @copper.hat I have seen it on two past quals form different schools, so no book reference unfortunately. May 8 '20 at 2:53

$$0 \le g_n - f_n \le h_n -f_n$$, 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 $$\varlimsup \int g_n \le \int g$$.

• 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? May 8 '20 at 3:00
• 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. May 8 '20 at 3:01
• For the DCT don't we need that the sequence is dominated by a fixed integrable function? May 8 '20 at 3:04
• @stochasticboy321 he probably means Generalized DCT in this case as i mention in my question. May 8 '20 at 3:04
• @Sorfosh Oh cool. I've either forgotten this or never read it, so thanks for the name of the theorem! May 8 '20 at 3:05