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In an article I'm currently reading, a reasoning is used that I don't understand.

We have an integral of a function over a domain with both depending on the same $\epsilon>0$. They show that $$\displaystyle\int_{D_\epsilon}f_\epsilon\,dx \to \displaystyle\int_{D}f\,dx \,\,\,\,\,\,\,\, (*)$$

as $\epsilon\to 0.$ There the integrand $f_\epsilon$ converges a.e. to $f$ and $\mathcal{L}^n(D_\epsilon\setminus D)$ vanishes. First they show of course that limit and integration can be changed in the first term. So far so good. Then however they say, that $(*)$ only holds for a subsequence and not the whole sequence. Of course it doesn't matter if we have to pass to a subsequence but I can't understand why that is? I thought about using the fact that $L^1$ convergence implies existence of a subsequence with convergence a.e. but I am not sure whether this is used or not here. Thanks in advance!

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    To understand why, add a well-suited "typewritter style" familly $g_\epsilon$ (see here) to $f_\epsilon$ (for a sequence, $g_n$ looks like a moving indicator function whose support goes to $0$ but moves around). If will not change the fact that (*) is true but a.e. convergence does not happen. – nicomezi Jan 05 '23 at 07:02
  • Thank you! That is a nice counterexample! – HelloEveryone Jan 05 '23 at 07:11

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