About definition of $f_n \rightarrow f$ in measure 
Definition: We say that $f_n \rightarrow f$ in measure to mean that for any $\epsilon >0, \mu(\{x: |f_n(x)-f(x)|> \epsilon \}) \rightarrow 0$ as $n \rightarrow \infty$.

My question for this is why is $\mu(\{x: |f_n(x)-f(x)|> \epsilon \})$ "$>$" not $<$? Thanks in advance.
 A: The concept of convergence of functions is that for every $x$ of interest, $|f_n(x) -f(x)|$ should go to $0$ as $n$ increases. That is, for every $\epsilon > 0$, $|f_n(x) - f(x)| < \epsilon$ for sufficiently large $n$. And it makes no difference it we replace that with $|f_n(x) - f(x)| \le \epsilon$, since a slightly smaller $\epsilon$ will make the stricter condition true.
For given $n, \epsilon$, the set $F_n^\epsilon = \{x : |f_n(x) - f(x)| > \epsilon\}$ is the set of all points where the condition $|f_n(x) - f(x)| \le \epsilon$ fails to hold. If $f_n \to f$ pointwise, then $F_n^\epsilon$ must eventually get smaller as $n$ increases, since for each $x$, there is some $N$ such that for $n > N, x \notin F_n^\epsilon$.
This is where the concept of convergence in measure comes from. Instead of demanding that at every point $f_n$ eventually gets close to $f$, it instead demands that the set of points where $f_n$ fails to be close to $f$ should shrink to insignficance as $n$ increases. For measure, "insignificance" means "measure $0$".
If you wanted to talk about $C_n^\epsilon = \{x : |f_n(x) - f(x)| < \epsilon\}$, the set where $f_n$ is close to $f$, the corresponding choice would be that $C_n^\epsilon$ should grow as $n$ increases until it is practically everything. So you might think about requiring $\mu\left(C_n^\epsilon\right) \to \mu(D)$, where $D$ is the domain of $f$ and the $f_n$. But there are pitfalls with this concept. In particular, it breaks down when $\mu(D) = \infty$. As then you can have $\mu\left(C_n^\epsilon\right) \to \mu(D)$, but still have the set where they fail also be infinitely large in measure.
So instead we define convergence in measure as we do, as it contains what is essential - that $f_n(x) \to f(x)$ except on insignificant subsets - without those problems.
