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I want to construct a sequence of function $\{f_n\}$ that does NOT converge point-wise in a domain $D$ but converges point-wise almost everywhere in $D$.

Define: $$f_n:D=[0,1]\cup\{2,3\}\to \mathbb R, ~~~\displaystyle f_n(x)=x^n,$$ Then $\{f_n\}$ does not converge point-wise as , when $x=2,3$ then $f_n\to \infty$ as $n\to \infty$. But for the function $$\displaystyle f(x)=\begin{cases}0 &\text{ if }0\le x<1\\1 &\text{ if }x=2,3\end{cases}$$ $f_n\to f$ point-wise almost everywhere in $D$ as , $m\left(\{2,3\}\right)=0$.

Is my construction correct ? If wrong please tell me where my fallacy and give a correct example.

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  • $\begingroup$ By some definitions that's not a domain because it isn't open. $\endgroup$ – Matt Samuel Mar 13 '16 at 13:45
  • $\begingroup$ Why bother considering extra points $2$ and $3$ ? The sequence $f_n:[0,1]\to\mathbb R, x\to x^n$ converges to the $0$ function almost everywhere. $\endgroup$ – Gabriel Romon Mar 13 '16 at 13:47
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    $\begingroup$ @LeGrandDODOM In your case, it does converge everywhere, though. Simply not to the zero function. (So you have to modify the example a bit, say $f_n\colon x\mapsto (-x)^n$.) $\endgroup$ – Clement C. Mar 13 '16 at 14:21
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Why not define $f_n:\mathbb{R}\to\mathbb{R}$ by $$f_n(x)=\left\{\begin{array}{ll} \frac{1}{n}&\mbox{ if }x\neq 0\\ (-1)^n&\mbox{ if }x=0 \end{array}\right.$$

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Seems correct to me. You could also consider $f_n=\sum_{k=0}^n x^k$ on $D=[0,1]$.

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