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I know that there are several questions about this topic, but my question is related to a specific argumentation.

Let $\Omega \subseteq \mathbb{R}^n$. We know that if a sequence of functions $f_n \to f$ in $L^p(\Omega)$, then there exists a subsequence $f_{n_k}$ (bounded by some $L^p$ function) which converges almost pointwise to $f$. In general $f_n$ does not converge pointwise to $f$, see e.g. the "type-writer sequence". However, what about the following argument?

If indeed $f_n$ does not converge pointwise to $f$, it means that there exists at least a subsequence $f_{n_m}$ which does not converge pointwise to $f$ i.e. there exist a set $A$ of non-zero measure and $\epsilon > 0$ such that for every $N >0$ there exists $m \ge N$ such that $$ |f_{n_m}(x)-f(x)| > \epsilon, \text{for every } x \in A \tag{A} \label{A} $$ On the other hand, $f_{n_m}\to f$ in $L^p$ (since $f_n$ converges in $L^p$, so any of its subsequences converges in $L^p$). Therefore, again, there exists a subsequence $f_{n_{m_l}}$ of $f_{n_m}$ which converges pointwise to $f$. But this last statement contradicts \eqref{A}.

What is the problem here? It might seem that the conclusion is "$f_n$ converges pointwise to $f$", but that would prove that any sequence converging in $L^p$ also converges pointwise, which is not true because of the counterexample of the typewriter sequence.

EDIT: \eqref{A} corrected according to the suggestion of @MotylaNogaTomkaMazura

moonknight
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  • (A) is not equivalent to $(f_{n_m})$ not converging a.e. to $f$. Even if you fix $x$, it is not equivalent to $(f_{n_m}(x))$ not converging to $f(x)$. – MaoWao Oct 26 '21 at 17:39
  • (A) does not follow. There is indeed a subsequence such that it does not hold. – Johan Aspegren Oct 26 '21 at 18:35
  • @JohanAspegren this is indeed what I meant when I wrote "this last statement contradicts (A)". But if (A) does not hold then $f_n$ must converge pointwise. But how is this possible if we know that the typewriter sequence converges in $L^p$ but not pointwise? – moonknight Oct 26 '21 at 18:52
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    @moonknight The sequence can oscillate pointwise: it comes infinitely often near the function value, but then goes further from the value. Especially the functions $f_n(x) $do not converge pointwise. – Johan Aspegren Oct 26 '21 at 19:00

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If $f_n$ does not converge to $f$ pointwise almost everywhere on $\Omega .$ Then for every $A\subset \Omega$ such that $\mu (A) =0 $ there exist $x_A\in \Omega \setminus A$ such that $f_n(x_A ) $ does not converge to $f (x_A ).$ This means that $$\forall_{A:\mu (A) =0}\exists_{x_A\in\Omega\setminus A}\exists_{\varepsilon >0}\forall_{n\in\mathbb{N}}\exists_{m\geq n} |f_m (x_A ) -f (x_A) |\geq \varepsilon$$ but the above condition does not imply existence a subseqence $f_{n_k} $ of $f_n$ such that $$|f_{n_k } (x) - f(x)|\geq \varepsilon $$ for all $k>N, $ where $\varepsilon >0 , N\in \mathbb{N}$ some fixed numbers.

  • You are right, thank you, I am going to correct my quantifiers. But still, my reasoning seems to contradict your condition as well, right? – moonknight Oct 26 '21 at 18:09