Convergence Theorems - Core Definitions

Understanding different modes of convergence for sequences of measurable functions is essential for analysis. Each type of convergence has distinct properties and applications.

DefinitionTypes of Convergence

Let $(X, \mathcal{F}, \mu)$ be a measure space and $\{f_n\}$ a sequence of measurable functions. We say $f_n$ converges to $f$ :

Pointwise: if $f_n(x) \to f(x)$ for all $x \in X$
Almost everywhere (a.e.): if $f_n(x) \to f(x)$ for almost every $x$ , i.e., $\mu(\{x : f_n(x) \not\to f(x)\}) = 0$
In measure: if for every $\epsilon > 0$ : $\lim_{n \to \infty} \mu(\{x : |f_n(x) - f(x)| \geq \epsilon\}) = 0$
In $L^p$ : if $\int_X |f_n - f|^p \, d\mu \to 0$ (for $1 \leq p < \infty$ )
Uniformly: if $\sup_{x \in X} |f_n(x) - f(x)| \to 0$

These modes of convergence form a hierarchy with different implications. Understanding their relationships is crucial for applying convergence theorems correctly.

ExampleImplications Between Convergences

The following implications hold (with caveats):

$\text{Uniform} \Rightarrow \text{Pointwise} \Rightarrow \text{Almost everywhere}$

$\text{Uniform} \Rightarrow L^p \Rightarrow \text{In measure}$

On finite measure spaces, convergence a.e. implies convergence in measure. However, convergence in measure does not imply a.e. convergence (though a subsequence converges a.e.).

For $L^p$ spaces, if $\mu(X) < \infty$ and $p \leq q$ , then $L^q$ convergence implies $L^p$ convergence.

Remark

Important observations:

A.e. vs. in measure: On $\mathbb{R}$ with Lebesgue measure, let $f_n = \chi_{[n, n+1]}$ . Then $f_n \to 0$ pointwise (hence a.e.), but NOT in measure, since $\lambda(\{x : |f_n(x)| \geq 1/2\}) = 1$ for all $n$ .
In measure vs. a.e.: Convergence in measure does not imply a.e. convergence. Consider the "typewriter sequence" on $[0,1]$ : $\chi_{[0,1]}$ , $\chi_{[0,1/2]}$ , $\chi_{[1/2,1]}$ , $\chi_{[0,1/3]}$ , $\chi_{[1/3,2/3]}$ , $\chi_{[2/3,1]}$ , etc. This converges to $0$ in measure but not pointwise anywhere.
$L^1$ vs. a.e.: $L^1$ convergence does not imply a.e. convergence, but combined with a bound on norms, it implies convergence in measure.