The Typewriter Sequence and Its Two Modes of Convergence

A Sequence That Moves Its Support

On a finite measure space, almost-everywhere convergence implies convergence in measure, but the converse need not hold. The dyadic typewriter sequence gives a concrete failure of that converse. Each term is an indicator of a short interval, and the lengths of those intervals tend to zero. Nevertheless, the intervals move through the unit interval in such a way that every point below one belongs to infinitely many of them and is excluded from infinitely many others.

Work on the finite complete measure space \(([0,1],\mathcal{M},m)\), where \(\mathcal{M}=\{E\cap[0,1]:E\in\mathcal{L}\}\) and \(m\) is Lebesgue measure restricted to \(\mathcal{M}\). All functions below are real-valued and measurable. For each integer \(k\ge1\), partition \([0,1)\) into the \(2^k\) half-open dyadic intervals

The point \(1\) is not in any of these intervals. We list the intervals level by level, starting with level \(k=1\). The indices assigned to level \(k\) are \(n=2^k-1+j\), with \(j=0,\ldots,2^k-1\). Define the sequence \((f_n)_{n\ge1}\) by

Thus each \(f_n\) takes only the values zero and one. In particular, every \(f_n\) is measurable, since its support is a Borel interval. The interval at level \(k\) has measure \(2^{-k}\); the important feature is that the sequence runs through every interval at that level before moving to the next one.

Reading the Enumeration

For convergence in measure to zero, the relevant sets are the error sets \(\{x:|f_n(x)|>\varepsilon\}\). Since each function is an indicator, their form depends on whether the threshold is below one or at least one. The strict inequality in this definition matters: when \(\varepsilon=1\), a value equal to one does not belong to the error set.

Convergence in Measure to Zero

Proof. Fix an arbitrary \(\varepsilon>0\), and let \(n=2^k-1+j\), where \(k\ge1\) is the level containing \(n\) and \(0\le j<2^k\). If \(0<\varepsilon<1\), then \(f_n\) equals one precisely on \(I_{k,j}\), and equals zero elsewhere. Therefore

If \(\varepsilon\ge1\), the error set is empty: on the support, \(|f_n(x)|=1\), which is not strictly greater than \(\varepsilon\), and off the support \(|f_n(x)|=0\). Its measure is then zero. It remains to check the limit as \(n\to\infty\). For every fixed \(K\ge1\), all indices in levels \(1,\ldots,K\) are at most \(2^{K+1}-2\), since level \(K\) ends at \(2^{K+1}-2\). Thus, once \(n>2^{K+1}-2\), its level satisfies \(k>K\). Consequently, the error-set measure in the case \(0<\varepsilon<1\) is at most \(2^{-(K+1)}\) for all such \(n\). Given any \(\delta>0\), choose \(K\) with \(2^{-(K+1)}<\delta\), and then take \(n>2^{K+1}-2\). The error-set measure is below \(\delta\). For \(\varepsilon\ge1\) it is always zero. This proves convergence in measure for every \(\varepsilon>0\). \(\square\)

Failure of Almost-Everywhere Convergence

Proof. Fix \(x\in[0,1)\). At each level \(k\), there is exactly one interval \(I_{k,j_k}\) containing \(x\), because the \(2^k\) half-open intervals are disjoint and their union is \([0,1)\). Equivalently, \(j_k=\lfloor 2^k x\rfloor\), where \(\lfloor y\rfloor\) is the greatest integer no larger than \(y\). At the index \(n_k=2^k-1+j_k\), the definition gives \(f_{n_k}(x)=1\). The indices \(n_k\) belong to successive, disjoint blocks, so they increase without bound.

At each level \(k\ge1\), there are \(2^k\ge2\) intervals. Choose an index \(j'_k\) in \(\{0,\ldots,2^k-1\}\) different from \(j_k\). The interval \(I_{k,j'_k}\) does not contain \(x\), so at the index \(n'_k=2^k-1+j'_k\), we have \(f_{n'_k}(x)=0\). These indices also increase without bound as the levels increase. Thus the values \(f_{n_k}(x)\) form a sequence constantly equal to one, while the values \(f_{n'_k}(x)\) form a sequence constantly equal to zero. If \(f_n(x)\) had a limit, both of these subsequences would have to tend to that same limit; their respective limits are one and zero. This is impossible. At \(x=1\), every support is contained in \([0,1)\), so \(f_n(1)=0\) for every \(n\). The set of points where convergence fails is therefore exactly \([0,1)\), whose measure is \(1-0=1\), not zero. Hence almost-everywhere convergence fails. \(\square\)

The Pitfall: Small Supports Do Not Force Pointwise Limits

A tempting but incorrect argument is: “The support measures \(m(\{f_n\ne0\})\) tend to zero, so \(f_n(x)\to0\) for every \(x\).” The measure calculation concerns the size of one support at a time; it does not say that a fixed point eventually leaves all supports. For a concrete correction, at \(x=1/3\) the terms \(f_1,f_4,f_9\) are all one, and the proof shows that this happens at an index in every level. At the same time, there are zero values at arbitrarily large indices. The shrinking supports are not nested: for instance, \(x=1/3\) lies in the support of \(f_4\), namely \([1/4,1/2)\), but not in the support of \(f_5\), namely \([1/2,3/4)\).

In contrast, convergence in measure compares the measure of the error set at one fixed threshold as the index grows. For \(0<\varepsilon<1\), that measure is \(2^{-k}\) throughout level \(k\), and \(2^{-k}\to0\) as \(k\to\infty\). These are different tests: one measures the size of individual error sets, while the other asks whether each point eventually has small errors. The earlier result that almost-everywhere convergence implies convergence in measure on finite measure spaces has no converse in general; this sequence demonstrates the failure on the finite interval \([0,1]\).