Theorems similar to Dini’s Theorem and Egoroff’s Theorem

Dini’s Theorem states that

Given a sequence of real-valued continuous functions (fn) on a compact set ER, if (fn) decreases to a continuous function f pointwise on E, then (fn) converges to f uniformly on E.

Egoroff’s Theorem states that

Given a measure space (E,A,μ) where ER has finite measure. A is an σ-algebra and μ is a measure on E.
If a sequence of measurable functions (fn) converges pointwise to f almost everywhere, then for every ε>0, there exists a measurable subset FE with μ(EF)<ε such that (fn) converges uniformly to f on F.

Question: Do there exist theorems, other than Dini and Egoroff, which give sufficient conditions for pointwise convergence to be uniform convergence?

I would like to see techniques involved in proving those theorems.
Dini and Egoroff used similar technique, which is to define a set containing elements such that fn and f are 'closed to each other'. If possible, I would like to know other technique to show pointwise convergence implies uniform convergence.


Here's one that I rather like that uses equicontinuity and compactness as the engine for moving from pointwise to uniform convergence. For the sake of clarity, let's recall what equicontinuity means.

Definition: Let X and Y be metric spaces. A set E of continuous functions from X to Y (i.e. EC(X,Y) ) is equicontinuous if for every xX and every ε>0 there exists δ>0 such that
yX and dX(x,y)<δdY(f(x),f(y))<ε for all fE.

With this definition in hand, we can state the result.

Theorem: Let X and Y be metric spaces with X compact. Suppose that for nN the function fn:XY is continuous and that {fn|nN} is equicontinuous. Let f:XY and suppose that fnf pointwise as n. Then fnf uniformly as n. In particular, this means that f is continuous.


Let ε>0. By the equicontinuity property, for each xX we can find δx>0 such that
yBX(x,δx)dY(fn(x),fn(y))<ε4 for all nN.
The collection of open balls {BX(x,δx)}xX is an open cover of X. The compactness of X allows us to find a finite subcover, namely x1,,xkX such that X=ki=1BX(xi,δxi). Due to the pointwise convergence fnf (or really the pointwise Cauchy property), for each i=1,,k we can choose NiN such that

Set N=max. For each x \in X there exists i\in \{1,\dotsc,k\} such that x \in B(x_i,\delta_{x_i}). Then for m \ge n \ge N we have the estimate

d_Y(f_m(x),f_n(x)) \le d_Y(f_m(x),f_m(x_i)) + d_Y(f_m(x_i),f_n(x_i)) + d_Y(f_n(x_i),f_n(x)) \\
< \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \frac{3\varepsilon}{4}.

Hence for n \ge N,

d_Y(f(x),f_n(x)) = \lim_{m \to \infty} d_Y(f_m(x),f_n(x)) \le \frac{3\varepsilon}{4}.

This holds for all x \in X, so

n \ge N \Rightarrow \sup_{x \in X} d_Y(f(x),f_n(x)) \le \frac{3\varepsilon}{4} < \varepsilon,

and we deduce that f_n \to f uniformly as n \to \infty.

Source : Link , Question Author : Idonknow , Answer Author : Glitch

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