How to show that the given sequence of functions converges to $f$ almost everywhere, almost uniformly and in measure?

Consider a sequence $(f_n)$ defined on $\mathbb{R}$ by $f_n =\chi_{[n,n+1]},$ $n\in \mathbb{N}$ and the function $f\equiv 0.$ Does $f_n$ converge to $f$ almost everywhere, almost uniformly or in measure?

$f_n\to f$ almost everywhere is the same as saying that $f_n\to f$ pointwise almost everywhere, i.e. on a subset whose complement has measure zero. Given $\epsilon>0$ and $x\in \mathbb{R}$ we observe that if $x\geq 0$ then for some $n_0\in \mathbb{N}$ we have that $n_0\leq xOn the other hand if $x<0$ then $f_{n}(x)=0$ and so we see that $$\lim_{n\to \infty}f_n(x) = f(x)$$ for all $x\in \mathbb{R}.$ Since the complement of $\mathbb{R}$ is $\emptyset$ which has measure $0$ we conclude that $f_n$ converges to $f$ almost everywhere.

Now $f_n$ does not converge uniformly to $f$ on the set $[0,\infty),$ a set of infinite measure. And so we conclude that $f_n$ does not converge almost uniformly to $f.$

If we choose $\epsilon = 0$ and $\eta =5$ then for all $N(\epsilon,\eta)\in \mathbb{N}$ we have that for $n\geq N.$
$$\mu(x\in D:|f_n|\geq \epsilon)=\mu(\mathbb{R})=+\infty\geq 5.$$

So we see that $f_n$ does not converge to $f$ in measure.

Is this solution correct?

This is problem 39 page 48 on the following pdf https://huynhcam.files.wordpress.com/2013/07/anhquangle-measure-and-integration-full-www-mathvn-com.pdf

The definition for convergence in measure used there is: