Measure convergence of $\frac{1}{n}$ 

**(1)**If $lim_{n\to\infty}\frac{1}{n}=0$, then why have $lim_{n\to\infty}\mu\{ x\in D:|f_n-f>0|\}=1$? Is this a proof by contradiction?
**(2)**Does(1) imply the condition (a) is true or false?
 A: The set $\{x \in D: |f_n(x) - f(x)| > 0\}$ is equal to the set $[0,1]$. So you have 
$$\lim_{n \to \infty} \mu([0,1]) = \lim_{n \to \infty} 1 = 1$$
1) is just saying that the converse of a) is not true. So a) is true, but its converse is false, as they show in 1)
EDIT
So, what they are saying is the following: We know that convergence in measure is defined as 
$$\forall \varepsilon > 0: \lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| \ge \varepsilon\} = 0$$
The more $\varepsilon $ is small, the harder it is to make the measure of that set "small". What happens then when $\varepsilon = 0$? In that case, the condition becomes 
$$\lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| > 0\} = 0$$
Intuitively, it seems that this condition is stronger than the previous one. It is true, and this is what they are saying. They are saying that


*

*The fact that $\lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| > 0\} = 0$ implies convergence in measure, i.e. implies that $\forall \varepsilon > 0: \lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| \ge \varepsilon\} = 0$. So the second condition implies the first.

*The fact that $\forall \varepsilon > 0: \lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| \ge \varepsilon\} = 0$ does not imply $\lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| > 0\} = 0$. 


So the condition is strictly stronger than what is requires for convergece in measure. How do they show this? With the following argument:
$f_n = 1/n$ is a sequence that converges almost uniformly to $f = 0$. Therefore, it must also converge in measure, therefore it must be that $\forall \varepsilon > 0: \lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| \ge \varepsilon\} = 0$. On the other hand, we can see how $\lim_{n \to \infty} \mu\{x \in D: |f_n(x) - f(x)| > 0\} = 1 \neq 0$. Therefore, it must be that convergence in measure does not imply the other condition, which is precisely what point 2) above is saying.
