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Let $\{f_n\}$ be a sequence of measurable functions on $E$.

Let $N_1<N_2<...$ such that $\forall i,j \geq N_n$ we have $m\{x \in E:|f_i(x)-f_j(x)| \geq \frac 1 {2^n}\}< \frac 1 {2^n}$. Oh, and also assume each $n<N_n.$

Now, I see that $\{f_{N_n}\}$ converges pointwise a.e. to a function $f$.

Let $\eta>0$.

Question: Can we show that $\limsup_{n \to \infty} m\{x \in E:|f_{N_n}(x)-f(x)|> \eta\}=0$?

My thought was to apply some sort of geometric series argument but, alas, after hours of work I still don't see how it can be done rigorously.

I see that it would be enough to show instead $\limsup_{n \to \infty} m\{x \in E:\sum_{i=n}^ \infty|f_{N_n}(x)-f(x)|> \eta\}=0$. Intuition suggests writing $\sum_{i=n}^ \infty|f_{N_n}(x)-f(x)|\leq \sum_{i=n} ^ \infty \frac 1 {2^i}$ but I think this move is illegal so I'm stuck.

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We know that $m\{x \in E: |f_{N_n}(x)-f_j(x)| >\eta\} <\frac 1 {2^{n}}$ if $j \geq N_n$ and $\eta >\frac 1 {2^{n}}$. Since $f_j \to f$ almost everywhere Fatou's Lemma this implies that $m\{x \in E: |f_{N_n}(x)-f(x)| >\eta\} \leq \frac 1 {2^{n}} $ if $\eta >\frac 1 {2^{n}}$. This finishes the proof.

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  • $\begingroup$ Would you mind explaining the Fatou's Lemma part? $\endgroup$ – Pascal's Wager Jan 16 '20 at 5:18
  • $\begingroup$ The set $\{x \in E: |f_{N_n}(x)-f(x)| >\eta\}$ is subset of $\lim \inf_j \{x \in E: |f_{N_n}(x)-f_j(x)| >\eta\}$ (ignoring a set of measure $0$). Fatou's Lemma says $m( \lim \inf A_j) \leq \lim \inf m(A_j)$. $\endgroup$ – Kavi Rama Murthy Jan 16 '20 at 5:23

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