# Do the assumptions give us $\limsup_{n \to \infty} m\{x \in E:|f_{N_n}(x)-f(x)|> \eta\}=0$?

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

Let $$N_1 such that $$\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

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.

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.
• 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 $m( \lim \inf A_j) \leq \lim \inf m(A_j)$. – Kavi Rama Murthy Jan 16 '20 at 5:23