real analysis - Uniform convergence and pointwise convergence, understanding

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?

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:

Fact: If a sequence of continuous functions $\{f_n\}$ converges pointwise to a function $f$, $f$ is not necessarily continuous.

---

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:

Lemma: If a sequence of continuous functions $\{f_n\}$ converges uniformly to a function $f$, $f$ is continuous.

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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:

Lemma: If a sequence of functions $\{f_n\}$ converges uniformly to a function $f$, $\{f_n\}$ converges pointwise almost everywhere.

Lemma: If a sequence of functions $\{f_n\}$, each of which can be identified with a continuous function, converges uniformly almost everywhere to a function $f$, $f$ can be identified with a continuous function.