Show that the sequence $\displaystyle f_n = n \chi_{[0, \frac{1}{n}]} \to 0$ almost everywhere.
- If $x < 0$, then $f_n(x) = f_{n+1}(x) =0.$
- If $0 \le x \le \frac{1}{n}$, then $f_n(x) = n < n+1 = f_{n+1}.$
- If $x> \frac{1}{n}$, then $f_n(x) = f_{n+1}(x) = 0.$
Notice that $f_n(x) \to 0$ everywhere except on the interval $[0, \frac{1}{n}]$.
We have $\lambda ([0, \frac{1}{n}]) = \frac{1}{n}$ for every $n$.
However $\frac{1}{n} \to 0$ as $n \to \infty$, i.e. $\lambda([0, \frac{1}{n}]) \to 0$ as $n \to \infty$.
That is, $f_n \to 0$ almost everywhere.
Is this correct?
$f_n\rightarrow 0$ almost everywhere is equivalent to say that the meausre of $E=\{x:lim_n f_n(x)\neq 0\}$ is zero, here $E=\{0\}$.