Convergence of functions

For measurable functions §

Related: Limits and accumulation points

Definition: Convergence for measurable functions

Let $f:E\to \mathbb{R}$ be a measurable function and let $f_n:E\to \mathbb{R}\cup \{\pm \infty \}$ be any function to the extended real numbers.

• $f_n\to f$ pointwise if for all $x\in E$ and all $ϵ>0$, there exists $N\in N$ such that

n > N \implies |f_n(x) - f(x)| < \epsilon.

- $f_n \to f$ **almost everywhere** if the set of all points which do not converge to $f$ has zero [[Lebesgue measure|measure]].

m({x \in E \ | \ f_n(x) \not \to f(x) }) = 0

- $f_n \to f$ **uniformly** if for all $\epsilon > 0$, there exists $N \in \mathbb N$ such that for all $x \in E$,

n > N \implies |f_n(x) - f(x) | < \epsilon.

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In metrizable (topological) spaces §

Uniformity of convergence for a sequence on functions depends on both the topology of their range and the metric.

Definition: Uniform convergence in metrizable spaces

Let $f_n:X\to Y$ be a sequence of functions from the set $X$ to the metric space $Y$. We say $(f_n)$ converges uniformly to $f:X\to Y$ if for all $ϵ>0$, there exists an integer $N$ such that for all $n\ge N$ and all $x\in X$, we have

$$d(f_n(x),f(x))<ϵ.$$

Abstract Topology HW 3.5: Relationship between uniform convergence and the uniform metric

Let $X$ be a set and let $f_n:X\to \mathbb{R}$ be a sequence of functions. Let $\overline{\rho }$ be the uniform metric on the space ${\mathbb{R}}^X$ of all functions $X\to \mathbb{R}$. If $f_n:X\to \mathbb{R}$ is a sequence of functions, show that $(f_n)$ converges uniformly to the function $f:X\to \mathbb{R}$ if and only if the sequence $(f_n)$ converges to $f$ as elements of the metric space $({\mathbb{R}}^X,\overline{\rho })$.