# The Typewriter Sequence

The typewriter sequence is an example of a sequence which converges to zero in measure but does not converge to zero a.e.

Could someone explain why it does not converge to zero a.e.?

$$f_n(x) = \mathbb 1_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} \text{, where } 2^k \leqslant n < 2^{k+1}.$$

Note: the typewriter sequence (Example 7).

I drew the first 63 functions in the sequence to help me understand its convergences. It might help others as well to understand answers given above:

Unfortunately, here I can only attach the rasterized format. In case someone wants to reproduce it, here is the Latex code:

\documentclass[9pt]{standalone}

\usepackage{bbm}
\usepackage{amsmath}
\usepackage{tikz,pgfplots}
\usetikzlibrary{arrows}

\newcommand{\nMAX}{63}
\newcommand{\xGrSamp}{8}

\begin{document}
\centering
\begin{tikzpicture}[font=\Large,shorten >=-2.5pt,shorten <=-2.5pt]
\begin{axis}[
axis x line*=bottom,
axis y line*=right,
axis z line*=left,
plot box ratio = 3 1000 2,
view={.3}{.2},
xmin=-0.2,    xmax=1.25,
ymin=0.6,    ymax=\nMAX+0.3,
zmin=0,    zmax=1.0,
xtick={0,1/8,2/8,3/8,4/8,5/8,6/8,7/8,1},
xticklabels={$0$,$\frac{1}{2^3}$,$\frac{2}{2^3}$,$\frac{3}{2^3}$,$\frac{4}{2^3}$,$\frac{5}{2^3}$,$\frac{6}{2^3}$,$\frac{7}{2^3}$,$1$},
ytick={0,...,\nMAX},
ztick={0,...,1.0},
xlabel=$x$,
ylabel=$n$,
zlabel=$f_n(x)$,
x label style={at={(axis description cs:0.067,-0.001)},anchor=north},
y label style={at={(axis description cs:0.062,0.145)},anchor=south},
z label style={at={(axis description cs:-0.002,0.035)},anchor=south},
yscale=5,
xscale=5,
legend entries={$f_n(x)=1\,$,
$f_n(x)=0\,$},
legend style={rounded corners=3pt,at={(0.023,0.14)}},
legend style={nodes={scale=1.5, transform shape}},
legend plot pos=right,
]
\foreach \n in {1, ..., \nMAX}
{
\pgfmathsetmacro\k{floor(log2(\n+1e-1))}
\pgfmathsetmacro{\xm}{-0.2}
\pgfmathsetmacro\xM{1.2}
\pgfmathsetmacro\xa{(\n-(2^(\k)))/(2^(\k))}
\pgfmathsetmacro\xb{(\n-(2^(\k))+1)/(2^(\k))}
\edef\temp
{
\noexpand\coordinate (d1) at (axis cs:\xm,\n,0);
\noexpand\coordinate (d2) at (axis cs:\xa,\n,0);
\noexpand\coordinate (d3) at (axis cs:\xa,\n,1);
\noexpand\coordinate (d4) at (axis cs:\xb,\n,1);
\noexpand\coordinate (d5) at (axis cs:\xb,\n,0);
\noexpand\coordinate (d6) at (axis cs:\xM,\n,0);
\noexpand\coordinate (g0) at (axis cs:\xm,\n,1);
\noexpand\coordinate (g1) at (axis cs:\xM,\n,1);
}
\temp
\draw[blue,<-o] (d1)--(d2);
\draw[black,dashed,line width=0.04mm] (d2)--(d3);
\draw[red,*-*]   (d3)--(d4);
\draw[black,dashed,line width=0.04mm] (d4)--(d5);
\draw[blue,o->] (d5)--(d6);
\draw[black,dashed,line width=0.04mm] (g0)--(g1);
}
\pgfplotsinvokeforeach{0, ..., \xGrSamp}
{
\draw[black,dashed,line width=0.06mm] (axis cs:#1/\xGrSamp,0,0)--(axis cs:#1/\xGrSamp,\nMAX,0);
}
\end{axis}
\node[rectangle,draw,rounded corners=3pt,text width=7.7cm] at (29.3,1.7)
{\huge Typewriter Sequence: \\$f_n(x)={\mathbbm{1}}_{[\frac{n-2^k}{2^k},\frac{n-2^k+1}{2^k}]},$\\
$\forall\, k\geq 0\,\, \&\,\, 2^k\leq n<2^{k+1}$};
\end{tikzpicture}

\end{document}


### Animated visual demonstration

As the sequence progresses, the indicator functions move from left to right over $$[0,1]$$ and then half in width, when they reach the end, and then repeat this ad infinitem. But then, each left-to-right run of a particular width must, at some point, have a value of $$1$$ for every $$x$$ in the interval. Therefore, throughout the infinitely many left-to-right runs, every $$x$$ attains a value of $$1$$ infinitely many times. But since $$x$$ is usually $$0$$, it must have no pointwise limit.

On the other hand, the width of the interval over which each function is $$1$$ decreases to $$0$$, and the functions are eventually infinitesimally thin spikes. So their 'size' (both in terms of $$L^1$$ norm and measure) converges to zero.

This also explains the origin of the sequence's name, as it looks like the a typewriter's carriage return jumping it back to where it started.

I wrote the sequence in Desmos by using $$\frac{\operatorname{sign}\left(x-c\right)+1}{2}=\textbf{1}_{(c,\infty)}$$, except at $$x=c$$, where it is $$0.5$$, and $$1-\operatorname{sign}(x-c)^2=\begin{cases}1,&x=c\\0,&\text{else}\end{cases}$$. This gives an expression for the functions in terms of the $$\operatorname{sign}(\cdot)$$ function, which can be interpreted by the program and can be simplified into the form in the animation above. A link to the project is available here.

• Great animation! I will link to this page every time I mention the "typewriter sequence" again. Commented Jan 23, 2020 at 15:03
• amazing work and comment thank @Jam ! Commented Jul 11, 2022 at 18:29
• @Jam Incredible work ! Thank a lot for your very clear answer. Commented Apr 21, 2023 at 14:24

Note that at any choice of $x$ and for any integer $N$, there is an $n>N$ with $f_n(x)=1$. So, the numerical sequence $f_n(x)$ cannot converge to $0$.

Note, however, that we can certainly select a subsequence of this sequence of functions that converges pointwise a.e.

• So $f_n(x)$ converges to 0, except for uncountable set? Does this set have measure > zero? or it is not measurable?
– MATH
Commented Aug 28, 2015 at 0:54
• @MATH I don't understand your question. For any $c \in [0,1]$, $f_n(c)$ is a non-convergent sequence over $n$. Does that make more sense? Commented Aug 28, 2015 at 3:26
• I mean it is asked to prove that it does not converge to zero a.e. but what you show is that it does not converge to zero?
– MATH
Commented Aug 28, 2015 at 10:03
• What I show is that it does not converge to zero anywhere (at any particular point). Certainly, then, it fails to "converge to $0$ a.e." (at almost every point). Commented Aug 28, 2015 at 10:06

Draw a picture of the generic function $f_n$ in the typewriter sequence. It's a rectangle of height 1 over an interval of width $1/2^k$, with value zero elsewhere. As the sequence progresses, the rectangles slide across the unit interval, the way a typewriter moves across the page. At each 'carriage return' of the typewriter, a new row of rectangles starts, each rectangle having half the width as before. You can see that for every point $x$ in the unit interval, the sequence $f_n(x)$ takes values zero and one infinitely often, so $f_n(x)$ cannot converge to any number.

I wrote this sequence of functions as a counterexample for a question that was deleted.

$$f_n(x)=\left[\frac km\le x\lt\frac{k+1}m\right]$$ where $$[\dots]$$ are Iverson brackets, $$m=\left\lfloor\frac{1+\sqrt{1+8n}}2\right\rfloor$$ and $$k=n-\frac{m(m-1)}2$$.

The Typewriter Sequence is a subsequence of this sequence (when $$m$$ is a power of $$2$$). Like the Typewriter Sequence, this sequence does not tend to $$0$$ pointwise, but its $$L^1$$ norms vanish.

The set on which $${f_n}$$ fails to converge to $$0$$ is exactly $$[0, 1]$$, which has full measure $$1$$. To see this, note that the interval $$[0, 1]$$ is repeatedly overwritten by indicator functions of intervals with decreasing length($$1/2$$, $$1/4$$, ... and so on). Since any point in $$[0, 1]$$is covered infinitely many times by such intervals, they oscillate between $$0$$ and $$1$$ infinitely often as $$n \to \infty$$.

The "overwriting" nature is what prevent the sequence from converging a.e, while the sequence $${{f_n := n 1_{[\frac{1}{n}, \frac{2}{n}]}}}$$ does converge to $$0$$ a.e.(actually it converges almost uniformly.)