A Sequence That Moves Rather Than Settles
A sequence of functions can become small in measure even though it does not settle down at almost any individual point. The typewriter sequence makes this contrast explicit: at each stage its value is one on a single interval, and the interval gets shorter at later scales. But the interval moves through the unit interval, so each point in \([0,1)\) is still visited repeatedly. The shrinking size controls convergence in measure; the repeated visits prevent pointwise convergence.
All statements below use the finite, complete measure space \(([0,1],\mathcal{L}_{[0,1]},m)\), where \(\mathcal{L}_{[0,1]}\) is the Lebesgue sigma-algebra restricted to \([0,1]\), and \(m\) is Lebesgue measure. The functions are real-valued and take only the values zero and one. Convergence in measure to zero means that, for every \(\eta>0\), \(m(\{x:|f_n(x)|>\eta\})\to0\). Convergence almost everywhere to zero means that there is a measurable set of measure zero outside which \(f_n(x)\to0\) pointwise. The strict inequality in the convergence-in-measure definition matters.
Defining the Typewriter Sequence
Thus the sequence is indexed starting at \(n=0\). More explicitly, within the block at scale \(r\), put \(k=n-(2^r-1)\). If \(k=0\), then \(f_n=\chi_{I_{r,2^r-1}}\). If \(1\le k\le2^r-1\), then \(f_n=\chi_{I_{r,k-1}}\). The intervals at scale \(r\) partition \([0,1)\), and each has measure \(2^{-r}\). They do not contain the endpoint \(1\), so every \(f_n(1)=0\).
Worked Example: Listing the First Blocks
At scale \(r=0\), the only interval is \(I_{0,0}=[0,1)\), so \(f_0=\chi_{[0,1)}\). At scale \(r=1\), the block starts at index \(2^1-1=1\), and the order is \(I_{1,1},I_{1,0}\). Hence \(f_1=\chi_{[1/2,1)}\) and \(f_2=\chi_{[0,1/2)}\).
At scale \(r=2\), the block starts at \(2^2-1=3\). The four intervals, in order, are \(I_{2,3},I_{2,0},I_{2,1},I_{2,2}\), giving \(f_3=\chi_{[3/4,1)}\), \(f_4=\chi_{[0,1/4)}\), \(f_5=\chi_{[1/4,1/2)}\), and \(f_6=\chi_{[1/2,3/4)}\). At scale \(r=3\), the block starts at \(2^3-1=7\), and its order is \(I_{3,7},I_{3,0},I_{3,1},I_{3,2},I_{3,3},I_{3,4},I_{3,5},I_{3,6}\). In particular, the last interval is at index \(7+7=14\), so \(f_{14}=\chi_{[3/4,7/8)}\).
Check: the scale-three block has indices \(7,8,9,10,11,12,13,14\), exactly eight indices for its eight intervals. Its last interval is \(I_{3,6}=[6/8,7/8)=[3/4,7/8)\), as required by the stated ordering.
Convergence in Measure
Proof. Fix \(\eta>0\). If \(0<\eta<1\), then \(f_n\) is the indicator of one interval \(I_{r,j}\) in its block, so \(\{x:|f_n(x)|>\eta\}=I_{r,j}\). Its measure is \(2^{-r}\). If \(\eta\ge1\), then \(|f_n(x)|\le1\) everywhere, so the strict-threshold set \(\{x:|f_n(x)|>\eta\}\) is empty. It remains to show that the scale \(r\) tends to infinity with \(n\). The block at scale \(r\) ends at \(2^{r+1}-2\), so for any integer \(R\ge0\), every \(n\ge2^R-1\) belongs to a block of scale \(r\ge R\). Given \(\delta>0\), choose \(R\) so that \(2^{-R}<\delta\). For \(0<\eta<1\) and \(n\ge2^R-1\), the threshold-set measure is \(2^{-r}\le2^{-R}<\delta\); for \(\eta\ge1\), it is zero. This is precisely convergence in measure to zero. \(\square\)
Worked Example: Checking a Threshold at a Specific Index
At index \(n=14\), the interval lies in the scale-three block, and the ordering gives \(f_{14}=\chi_{I_{3,6}}=\chi_{[3/4,7/8)}\). Take the threshold \(\eta=1/2\). Since \(f_{14}\) is either zero or one, \(|f_{14}(x)|>1/2\) holds exactly on \([3/4,7/8)\). Thus
For the same function and threshold \(\eta=1\), the strict inequality \(|f_{14}(x)|>1\) holds nowhere, so the threshold-set measure is zero. Check: the interval length is \(1/8\), which equals \(2^{-3}\), the length prescribed by its scale.
Why Almost-Everywhere Convergence Fails
Proof. Fix \(x\in[0,1)\). At each scale \(r\), the intervals \(I_{r,0},\ldots,I_{r,2^r-1}\) partition \([0,1)\), including their left endpoints and excluding their right endpoints. Therefore there is exactly one index \(j\) with \(x\in I_{r,j}\). In the scale-\(r\) block, the function equal to the indicator of this interval takes value one at \(x\). Since there is one such index in every block and there are infinitely many blocks, \(f_n(x)=1\) for infinitely many \(n\).
For every \(r\ge1\), the block contains \(2^r\ge2\) intervals. Only one of them contains \(x\), so at least one other interval does not contain \(x\); the function at that other index has value zero at \(x\). Since this holds in every block with \(r\ge1\), \(f_n(x)=0\) for infinitely many \(n\) as well. A real sequence taking both zero and one infinitely often cannot converge: if it converged to \(L\), its subsequence constantly equal to zero would force \(L=0\), while its subsequence constantly equal to one would force \(L=1\). This contradiction proves failure of pointwise convergence at every \(x\in[0,1)\). The remaining point \(x=1\) has measure zero, so failure on the measure-one set \([0,1)\) rules out almost-everywhere convergence. \(\square\)
Worked Example: Following the Point x = 3/4
The endpoint \(3/4\) belongs to intervals whose left endpoint is \(3/4\), because the dyadic intervals are closed on the left. At scale \(r=1\), \(3/4\in I_{1,1}=[1/2,1)\), so \(f_1(3/4)=1\). At scale \(r=2\), \(3/4\in I_{2,3}=[3/4,1)\), the first interval in that block, so \(f_3(3/4)=1\).
At scale \(r=3\), \(3/4\in I_{3,6}=[3/4,7/8)\). This is the eighth interval listed in that block: \(I_{3,7}\) is first, followed by \(I_{3,0},\ldots,I_{3,6}\). The block starts at index \(7\), so the eighth entry has index \(7+7=14\), and \(f_{14}(3/4)=1\). At the same scale, \(f_7\) is the indicator of \(I_{3,7}=[7/8,1)\), which does not contain \(3/4\), so \(f_7(3/4)=0\).
Check: the values at indices \(1,3,7,14\) include \(1,1,0,1\), respectively. The one at index \(14\) follows from the actual block order, not from treating \(I_{3,6}\) as an early entry.
A Subsequence Can Behave Differently
Failure of almost-everywhere convergence for the full sequence does not mean that every subsequence fails to converge almost everywhere. For example, select the first index in each block, \(n_r=2^r-1\). By the definition of the ordering, \(f_{n_r}=\chi_{I_{r,2^r-1}}=\chi_{[1-2^{-r},1)}\). These intervals have measures \(2^{-r}\), which tend to zero. In fact this subsequence converges pointwise to zero everywhere on \([0,1]\).
Worked Example: Verifying a Convergent Subsequence
Let \(x<1\). Then \(1-x>0\), so choose \(R\) such that \(2^{-R}<1-x\). For every \(r\ge R\), the left endpoint of the selected interval satisfies \(1-2^{-r}\ge1-2^{-R}>x\). Hence \(x\notin[1-2^{-r},1)\), and \(f_{n_r}(x)=0\) for all \(r\ge R\). At \(x=1\), the interval excludes its right endpoint, so \(f_{n_r}(1)=0\) for every \(r\). This proves pointwise convergence to zero on all of \([0,1]\).
The integral of each selected indicator is the measure of its interval: \(\int f_{n_r}\,dm=2^{-r}\). For instance, at \(r=1\), \(n_r=1\), the interval is \([1/2,1)\), and its measure is \(1-1/2=1/2=2^{-1}\); at \(r=2\), \(n_r=3\), the interval is \([3/4,1)\), and its measure is \(1-3/4=1/4=2^{-2}\). In general the length is \(1-(1-2^{-r})=2^{-r}\), which tends to zero. The pointwise conclusion and the integral calculation agree with the shrinking supports.
The Pitfall: Shrinking Supports Do Not Stay Put
A tempting but incorrect argument is: “The support intervals have lengths tending to zero, so each point is eventually outside them; therefore the sequence converges pointwise to zero.” The implication from short intervals to eventual exclusion is false when the intervals move. For the concrete point \(x=3/4\), the scale-one, scale-two, and scale-three blocks each include an interval containing \(x\), at indices \(1,3,14\), respectively. The same argument applies at every scale, so this point is included infinitely often, despite the interval lengths \(1/2,1/4,1/8,\ldots\) tending to zero.
The distinction is between measuring a set of exceptional points at one index and tracking a fixed point over all indices. Convergence in measure asks whether the measure of the threshold set is small for each sufficiently late function. Almost-everywhere convergence asks whether, outside one null set, each individual point eventually has values close to the limit. For the typewriter sequence, each individual support is small at late indices, but the supports cycle so that no point of \([0,1)\) is eventually avoided.
Check Your Understanding
Use the block definition and keep the half-open endpoints and strict threshold inequality explicit.
- Which interval and index correspond to the fourth entry of the scale-three block, and what is the value of the corresponding function at \(x=1/2\)?
- For a scale-\(r\) function and \(0<\eta<1\), compute \(m(\{x:|f_n(x)|>\eta\})\). How does the answer change when \(\eta=1\)?
- Explain why every \(x\in[0,1)\), including dyadic endpoints, belongs to exactly one interval at each scale.
- Why does having both zero and one occur infinitely often prevent convergence of the real-valued sequence \(f_n(x)\)?
- For the subsequence \(n_r=2^r-1\), identify its supporting interval and show why a fixed \(x<1\) eventually lies outside it.