Showing that a sequence of piecewise functions converges 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!
 A: 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 $x$).
