I already know that $\chi_n := \chi_{(n,\infty)}$ converges (almost) everywhere to $0$, and I want to see if it converges almost uniformly too.
Fix $\varepsilon > 0$. It seems to me that $\chi_n$ does not converge in (Lebesgue) measure, because for any $N>0$, $n>N$ implies $\mu \{x \mid |\chi_n(x)| = 1\} > 0$. So by a theorem saying "convergence almost uniformly implies convergence in measure", the sequence $\{ \chi_n \}$ does not converge almost uniformly. Is this ok?
I also want to check the definition of almost uniform convergence too, and find a contradiction: if there were a set $E \subset \bf{R}$ such that $\mu(E) < \varepsilon$ and $\chi_n$ converging uniformly on $\mathbf{R} \setminus E$, then we require the existence of $N>0$ such that $n>N$ implies $\chi_n(x) = 0$ for all $x \in \mathbf{R} \setminus E$. I assume $E \subset (n, \infty)$ for otherwise the $E$ does not affect $\chi_n$ and $\chi_n$ does not converge almost uniformly. In some way I want to say that $(n, \infty) \setminus E$ still has nontrivial measure and hence no $N$ can ensure the almost uniform convergence. But I don't know how to formalise this (if it's even a good idea). Thanks in advance.
Your argument using convergence in measure is correct.
For the direct part, assume that $\left(\chi_n\right)_{n\geqslant 1}$ converges uniformly on some set $F$. Then $\sup_{x\in F}\left\lvert\chi_n(x)-\chi_{n+1}(x)\right\rvert\to 0 $ and since $\left\lvert\chi_n(x)-\chi_{n+1}(x)\right\rvert$ can take only the values $0$ and $1$, we establish the existence of an integer $N$ such that $\sup_{x\in F}\left\lvert\chi_n(x)-\chi_{n+1}(x)\right\rvert=0$ for $n\geqslant N$. Therefore, for each $n\geqslant N$, $F\cap \left[n,n+1\right)=\emptyset$. This implies that the measure of the complement of $F$ is bigger than one (actually infinite) hence almost uniform convergence is not possible.