Subtle difference between convergence in measure and almost uniform convergence? I am reading Terence Tao's post on convergence types.
Consider the following two types:

We say a sequence $f_n : \mathbb R \to \mathbb R$ converges in measure
  to a function $f$ if and only if for every $\varepsilon > 0$ 
$$\mu(\{x \in \mathbb R \mid |f_n(x) - f(x)|\ge \varepsilon \})\to 0$$
as $n \to \infty$.

On the other hand, 

$f_n $ converges to $f$ almost uniformly if for every $\varepsilon >
 0$ there is a set $E$ such that $\mu(E) \le \varepsilon$ and outside
  $E$, $f_n\to f$ uniformly.

To me these seem very similar. Concretely, the expression $\mu(\{x \in \mathbb R \mid |f_n(x) - f(x)|\ge \varepsilon \})\to 0$ could be rewritten to say that given any $\delta > 0$ then $\mu(\{x \in \mathbb R \mid |f_n(x) - f(x)|\ge \varepsilon \}) \le \delta$. Then the set $\{x \in \mathbb R \mid |f_n(x) - f(x)|\ge \varepsilon \}$ would be the set $E$ outside of which $|f_n - f |<\varepsilon$ that is, outside of which $f_n \to f$ uniformly.

Question Please could someone elaborate on the difference of these two
  definitions? Perhaps with an example? I need some help gaining
  intuition.

 A: The main difference is that or almost uniform convergence, you fix the exceptional set $E$ with $\mu(E)<\gamma$ once and for all (only depending on $\gamma$) and then you want $f_n \to f$ on uniformly on $E^c$. Note that $E$ is not allowed to change with $n$ or with the $\epsilon$ distance that we choose when we want to exploit the uniform convergence. 
In contrast, if we have convergence in measure, then the set
$$
E_{n,\epsilon}=\{x \mid |f(x)-f_n (x)|>\epsilon\}
$$
of which we know that it has small measure for $n \to \infty$ depends on $n$ and $\epsilon$.
Thus, it could in principle happen (and there are such examples) that we have 
$$
\Omega = \bigcup_{n\geq N} E_{n,\epsilon}
$$
for all sufficiently small $\epsilon$ and all $N$, i.e. the union of "problematic sets" still covers all of $\Omega$. 
One example where this happens is $\Omega=[0,1]$ and for the sequence
of indicator functions of the sets
$$
[0,1],[0,1/2],[1/2,1],[0,1/3],[1/3,2/3],[2/3,1],[0,1/4],\dots. 
$$
Having said that, a remarkable theorem of Egoroff shows that if $\Omega$ has finite measure and if $(f_n)_n$ converges in measure, we can always choose a subsequence which converges almost uniformly. 
