Sheng Li (gmachine1729) wrote,
Sheng Li
gmachine1729

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.

 

Tags: analysis, measure theory
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