By definition, I worked out that

$$ \require{cancel} \{\{f_n\}, f : f_n \overset{a.e.}{\to} f\} = \left\{\{f_n\}, f : m\left(\bigcup_{k=1}^{\infty} \bigcap_{n_k = 1}^{\infty} \bigcup_{n_k}^{\infty} \{x \in E : |f_k(x) - f(x)| \ge \frac{1}{k}\}\right) = 0 \right\} $$

To elobrate, this is because $\bigcup_{k=1}^{\infty} \bigcap_{n_k = 1}^{\infty} \bigcup_{n_k}^{\infty} \{x \in E:|f_k(x) - f(x)| \ge \frac{1}{k}\}$ is the defintion of $\{x \in E:f_n \cancel{\to} f\}$. For convergence in measure, there is also

$$ \{\{f_n\}, f : f_n \Rightarrow f\} = \bigcap_{i=1}^{\infty} \bigcap_{j=1}^{\infty} \bigcup_{n_k=1}^{\infty} \bigcap_{k=n_k}^{\infty} \left\{\{f_n\}, f : m\left(x \in E: |f_k(x) - f(x)| \ge \frac{1}{i} \right) < \frac{1}{j} \right\} $$

How do you express uniform convergence almost everywhere with this set operations like this?


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