Why does $f_n \to 0$ almost everywhere?

Question

Let $\displaystyle f_n = n \chi_{[\frac{1}{n}, \frac{2}{n}]}$ for all $n \in \mathbb N$, then $f_n \to 0$ almost everywhere.

Let $x \in \mathbb R$, then there exists some $N \in \mathbb N$ such that $x \not\in [\frac{1}{n},\frac{2}{n}]$ for all $n \ge N$.

Then $f_n(x) = 0$, provided that $n \ge N$.

Why does this imply that $f_n \to 0$ almost everywhere? I cannot seem to see how this satisfies the definition of "almost everywhere", i.e. the set where $f_n \not\to 0$ mus have measure 0.

Answer 1

You've shown that the set of $x$ where $(f_n(x))$ does not converge to $0$ is empty. Of course, the empty set has measure zero.

Answer 2

Actually $f_n \rightarrow 0 $ everywhere but in maths it's usually not used - convergency 'almost everywhere' is enough