Why $\frac{1}{n}\mathcal{X}_{[0,n]}$ converges to zero in measure?

Question

How can I directly prove that for any $\epsilon>0, \lim_n m\left\{x:|f_n(x)|>\epsilon\right\}=0?$

I have a confusion. Let $\epsilon=1/2$ then $\left\{x:f_n(x)>1/2\right\}=[0,1]$ but $m[0,1]\not<1/2$

Let $\epsilon>0$ then $\left\{x:f_n(x)>\epsilon\right\}=[0,n]$ with $1/n>\epsilon$ then $m([0,n]\not<\epsilon$... It's wrong?

Answer 1

Given $\epsilon>0$ you can find $N$ such that $N<n$ implies $1/n < \epsilon$, so the set you want to measure is empty.

Answer 2

Let's look at the values of $f_{n}(x) := \frac{1}{n} \chi_{[0,n]}(x)$. This function can be equivalently expressed as $$f_{n}(x) = \begin{cases} \frac{1}{n} & x \in [0,n] \\ 0 & \text{else} \end{cases}.$$

So, $f_{1}(x) = 1$ on $[0,1]$ and $0$ everywhere else. $f_{2}(x) = \frac{1}{2}$ on $[0,2]$ and $0$ everywhere else. $f_{10}(x) = \frac{1}{10}$ on $[0,10]$ and $0$ everywhere else.

Clearly, $|f_{n}(x)| \leq \frac{1}{n}$ for every $x$. Now, given any $\epsilon > 0$, we can find $m \in \Bbb N$ so that $\frac{1}{m} \leq \epsilon$, right? Thus, $|f_{m}(x)| \leq \frac{1}{m} \leq \epsilon$. Also, if $n \geq m$, then we also have $\frac{1}{n} \leq \frac{1}{m} \leq \epsilon$.

What does that tell you about $|f_{n}(x)|$ in relation to $\epsilon$ for every $n \geq m$? What does this imply about the set $\{x \mid |f_{n}(x)| > \epsilon \}$ for every $n \geq m$?