# Convergence in measure

Originally published at 狗和留美者不得入内. You can comment here or there.

Let $f, f_n (n \in \mathbb{N}) : X \to \mathbb{R}$ be measurable functions on measure space $(X, \Sigma, \mu)$. $f_n$ converges to $f$ globally in measure if for every $\epsilon > 0$, $\displaystyle\lim_{n \to \infty} \mu(\{x \in X : |f_n(x) - f(x)| \geq \epsilon\}) = 0$.

To see that this means the existence of a subsequence with pointwise convergence almost everywhere, let $n_k$ be such that for $n > n_k$, $\mu(\{x \in X : |f_{n_k}(x) - f(x)| \geq \frac{1}{k}\}) < \frac{1}{k}$, with $n_k$ increasing. (We invoke the definition of limit here.) If we do not have pointwise convergence almost everywhere, there must be some $\epsilon$ such that there are infinitely many $n_k$ such that $\mu(\{x \in X : |f_{n_k}(x) - f(x)| \geq \epsilon\}) \geq \epsilon$. There is no such $\epsilon$ for the subsequence $\{n_k\}$ as $\frac{1}{k} \to 0$.

This naturally extends to every subsequence’s having a subsequence with pointwise convergence almost everywhere (limit of subsequence is same as limit of sequence, provided limit exists). To prove the converse, suppose by contradiction, that the set of $x \in X$, for which there are infinitely many $n$ such that $|f_n(x) - f(x)| \geq \epsilon$ for some $\epsilon > 0$ has positive measure. Then, there must be infinitely many $n$ such that $|f_n(x) - f(x)| \geq \epsilon$ is satisfied by a positive measure set. (If not, we would have a countable set in $\mathbb{N} \times X$ for bad points, whereas there are uncountably many with infinitely bad points.) From this, we have a subsequence without a pointwise convergent subsequence.

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