Does $f_n(x) = \chi_{(n, n+1)}$ converge uniformly on $\mathbb{R}$ Does $f_n(x) = \chi_{(n, n+1)}$ converge uniformly to the zero function on $\mathbb{R}$?
It seems to me that it doesn't.
Say $\exists$ $N$ such that for all $n \ge N$
$$|\chi_{(n, n+1)} - 0|< \epsilon$$ with $\epsilon > 0$
i.e. $|\chi_{(n, n+1)}|< \epsilon$
In particular,
$|\chi_{(N, N+1)}|< \epsilon$
Well, if $N < x < N + 1$, then $f_n(x) = 1 \not< \epsilon$
So $f_n(x)$ is not uniformly convergent, it's rate of convergence is dependent on $x$.
Have I got all that correct? If so, how do I interpret the dependence on $x$? Is it that if $x$ is very large it takes longer to converge to the zero function than when $x$ is very small?

You can prove this by the definition.
For $\epsilon_0=\frac12$, for any $N\in \mathbb{N}$, choose $n_0=N+1>N $ and $x_0=N+\frac32$, then $|f_{n_0}(x_0)|=|\chi_{(N+1, N+2)}(N+\frac32)|=1>\epsilon_0$. This means that $f_n$ does not uniformly converges on $\mathbb{R}$.
