Pointwise convergence $\ \ \Leftrightarrow \ \ \{f_n(x)\}_{1}^{\infty}$ converges to $f(x)$ for all $x\in X$
Uniform convergence $ \ \ \Leftrightarrow \ \ \forall \epsilon > 0 \exists N\in\mathbb{N} \forall x\in X \forall n\geq N \ \ s.t. \ \ |f_n(x) – f(x)| < \epsilon$
Pointwise almost everywhere convergence $ \ \Leftrightarrow \ \exists E\in M \ \ \text{with} \ \ \mu(E) = 0 \forall x\notin E \ \ f_n(x)\rightarrow f(x)$
In class my professor stated that uniform implies pointwise which implies pointwise almost everywhere. Can anyone explain why this is the case, what is the real difference between uniform and pointwise convergence, and is there a proof to show this?
Best Answer
The problem with pointwise convergence
Pointwise convergence does not generally preserve some properties of functions that you would expect to be preserved in limits. For example, consider the functions $f_{n}(x)=x^{n}$ defined on $[0,1]$.
This sequence of functions has a pointwise limit: $$f(x)=\lim_{n\rightarrow\infty}f_{n}(x)=\lim_{n\rightarrow\infty}x^{n}=0 \text{ if } 0\leq x<1 \text{ and }f(1)=\lim_{n\rightarrow\infty}f_{n}(1)=\lim_{n\rightarrow\infty}1^{n}=1.$$ Therefore, $$ f(x)=\begin{cases} 0 & \text{if }0\leq x<1;\\ 1 & \text{if }x=1. \end{cases} $$
Most alarming is the fact that $f_{n}$ is continuous for each $n$... but $f$ is not! To preserve important properties like continuity, we need to impose stronger conditions on convergence; that's where uniform convergence comes in. Let's summarize:
Uniform convergence
Suppose $f_{n}$ converges uniformly to $f$ and that $f_{n}$ is continuous for each $n$. Let $y$ be a point in the domain of $f$. Let $\epsilon>0$. By the assumption, we know that (i) there exists $N$ such that for all $n\geq N$ and for all $x$, $|f_{n}(x)-f(x)|<\epsilon/3$ and (ii) for each $n$, there exists $\delta_{n}>0$ such that for all $z$ satisfying $|y-z|<\delta_{n}$, $|f_{n}(y)-f_{n}(z)|<\epsilon/3$. Therefore, \begin{align*} |f(y)-f(z)| & =|f(y)-f(z)+f_{n}(y)-f_{n}(y)+f_{n}(z)-f_{n}(z)|\\ & =|(f(y)-f_{n}(y))-(f(z)-f_{n}(z))+f_{n}(y)-f_{n}(z)|\\ & \leq|f(y)-f_{n}(y)|+|f(z)-f_{n}(z)|+|f_{n}(y)-f_{n}(z)|\\ & <\epsilon/3+\epsilon/3+\epsilon/3=\epsilon \end{align*} so that $f$ is continuous! Let's summarize:
Measure-theoretic extension
Instead of identifying functions by their values on all points in a domain, in measure theory, we identify two functions when they agree almost everywhere. For example, under the usual notion of Lebesgue measure, the functions $$f(x)=x \text{ and } f(x)=\begin{cases} x & \text{if }x\neq1\\ 0 & \text{if }x=1 \end{cases}$$ are identical. It makes sense to relax our definition of convergence to take this into account. I will let you prove to yourself the following two facts: