Let $f_{k}:\mathbb{R}\rightarrow \mathbb{R}$ be given by
$$f_k(x)=\begin{cases} -k & \text{if } x<-k \\[6pt] x & \text{if } -k\leq x<k \\[6pt] k & \text{if } x>k \end{cases}$$
Show that the sequence $f_{k}$ converges pointwise, but there is no set $E$ with $|\mathbb{R}\setminus E|<\infty$ so that the sequence converges uniformly on $E$.
I'm thinking that the limiting function is $f_{k}(x)$? For example in the case that $x<-k$, $f_{k}(x)=-k$. Take any $n\in \mathbb{Z}^{+}$. Then for $k\geq n,$ $|f_{k}(x)-(-k)|=0<\epsilon$. I think the same should follow in the other two cases. Am I understanding pointwise convergence correctly? I'm not sure how to proceed with the second part. Any hints or guidance is welcome :) Thanks!
The limit of anything, as $k$ approaches something, does not depend on $k,$ so it can't be $f_k(x).$ You have a sequence $f_1,f_2,f_3,\ldots$. In the first term, $k$ is $1,$ in the second term, $k$ is $2,$ and so on. What would $k$ be in $f_k$ if $f_k$ is the limit?
Imagine some fixed value of $x.$ Take $x=8.3,$ for example. The sequence is \begin{align} & f_1(8.3) = 1 \text{ since } 8.3>1, \text{ i.e. } x>k \\ & f_2(8.3) = 2 \text{ since } 8.3>2 \\ & f_3(8.3) = 3 \text{ since } 8.3>3 \\ & f_4(8.3) = 4 \text{ since } 8.3>4 \\ & f_5(8.3) = 5 \text{ since } 8.3>5 \\ & f_6(8.3) = 6 \text{ since } 8.3>6 \\ & f_7(8.3) = 7 \text{ since } 8.3>7 \\ & f_8(8.3) = 8 \text{ since } 8.3>8 \\ & f_9(8.3) = 8.3 \text{ since } 8.3<9 \text{ i.e. } x<k \\ & f_{10}(8.3) = 8.3 \text{ since } 8.3<10 \\ & f_{11}(8.3) = 8.3 \text{ since } 8.3<11 \\ & \qquad\vdots \\ & \qquad \vdots \end{align} So when $x=8.3$ then the sequence is $$ 1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,8.3,\,8.3,\,8.3,\,8.3,\, \ldots $$ The limit is $8.3,$ i.e. the limit is $x.$
This sequence cannot converge uniformly on any unbounded set. For example, suppose we seek some index $k$ beyond which every term of the sequence is within $\varepsilon = \tfrac 1 {10}$ of the limit. Just make $x>(2+\text{that value of }k)$ and it fails to come that close to the limit (which is $x$).