In the top answer given to this question, the conclusion is that $$\mu\Big(\bigcap_{j=1}^\infty\bigcup_{k\geq j}A_k\Big)\geq \delta$$ But I don't understand why this implies that it doesn't converge pointwise. I was under the impression that we had to conclude $\lim f_i(x)\neq f(x)$. Can someone clear this up?
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This answer merely expounds upon angryavian's answer to answer D. Brio's question in the comments.
Suppose $x\in \bigcap_{j=1}^\infty \bigcup_{k=j}^\infty A_k$.
Then for any positive integer $N$, $\,x\in \bigcap_{j=N}^\infty \bigcup_{k=j}^\infty A_k$.
But then $x\in A_k$ for some $k\geq N$.
And that implies $x\in \{z | f_{m_k}(z) −f(z)|>\epsilon\}$ for some $k\geq N$.
Hence, $|f_{m_k}(x)-f(x)| > \epsilon$ for some $k\geq N$.
We have shown that for every positive integer $N$, there exists a $k\geq N$ such that $|f_{m_k}(x)-f(x)| > \epsilon$. So the $f_{m_k}(x)$ do not converge pointwise to $f(x)$ for any $x\in \bigcap_{k=1}^\infty \bigcup_{k=j}^\infty A_k$.
irchans
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