Three Modes, Three Different Questions
For a sequence of measurable functions, pointwise convergence asks what happens at each individual point; almost-everywhere convergence permits exceptions on a null set; convergence in measure asks whether the sets on which the error exceeds a fixed threshold become small in measure. These questions can give different answers even for simple indicator functions. A reliable comparison begins by fixing the candidate limit and calculating the error sets, with the strict inequality in their definition kept in view.
For the statements and examples below, functions are real-valued and measurable, and the candidate limit is real-valued and measurable. On a measure space \((X,\mathcal{M},\mu)\), convergence \(f_n\to f\) pointwise means \(f_n(x)\to f(x)\) for every \(x\in X\). Convergence \(f_n\to f\) \(\mu\)-a.e. means pointwise convergence outside some \(N\in\mathcal{M}\) with \(\mu(N)=0\). Convergence in measure means that for every \(\varepsilon>0\), the measurable error sets have measures tending to zero:
The threshold is strict: the error set is \(\{|f_n-f|>\varepsilon\}\), not \(\{|f_n-f|\ge\varepsilon\}\). Convergence in measure is defined this way on finite and infinite measure spaces alike; finiteness is not part of its definition. In each example, the underlying space and measure will be stated explicitly.
This gives a useful direct test for indicators. For any measure space \((X,\mathcal{M},\mu)\), with no finiteness assumption, \(\chi_{A_n}\to\chi_A\) in measure if and only if \(\mu(A_n\,\triangle\,A)\to0\). The pointwise and almost-everywhere versions instead ask whether membership in \(A_n\) eventually agrees with membership in \(A\).
Proof. If \(0<\varepsilon<1\), then \(|\chi_{A_n}(x)-\chi_A(x)|\) is either \(0\) or \(1\), and it equals \(1\) exactly when \(x\in A_n\,\triangle\,A\). Thus the error-set measure is precisely \(\mu(A_n\,\triangle\,A)\). If the indicators converge in measure, applying the definition at \(\varepsilon=1/2\) gives \(\mu(A_n\,\triangle\,A)\to0\). Conversely, if those symmetric-difference measures tend to zero, then for every \(0<\varepsilon<1\) the error-set measure tends to zero by the identity, while for \(\varepsilon\ge1\) the error sets are empty. This proves the in-measure equivalence. At an individual point, the indicator values lie in \(\{0,1\}\); convergence to \(\chi_A(x)\in\{0,1\}\) is equivalent to eventually having exactly that value, because a different value is at distance \(1\). The same pointwise statement outside a null set gives the almost-everywhere version. \(\square\)
Worked Comparisons
Worked Example: Indicators of Shrinking Intervals
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra restricted to \([0,1]\), and let \(\mu\) be Lebesgue measure on this finite measure space. Set \(A_n=[0,1/n]\) for \(n\ge1\) and \(f_n=\chi_{A_n}\). Determine the pointwise limit and calculate the error sets relative to it.
At \(x=0\), \(f_n(0)=1\) for every \(n\). If \(x>0\), then for every integer \(n>1/x\) we have \(1/n<x\), so \(x\notin A_n\) and \(f_n(x)=0\) thereafter. Consequently \(f_n\to f=\chi_{\{0\}}\) at every point of \([0,1]\), including \(0\). For \(0<\varepsilon<1\), the difference from the limit equals \(1\) precisely when \(x\in(0,1/n]\): at \(0\) both functions equal \(1\), and outside \([0,1/n]\) both equal \(0\). Therefore
For \(\varepsilon\ge1\), the error set is empty because the error never exceeds \(1\). Thus its measure tends to zero for every positive threshold, and \(f_n\to f\) in measure as well as pointwise and almost everywhere. Check: at \(n=4\), the error interval is \((0,1/4]\), whose measure is \(1/4\), matching \(1/n=1/4\). The value at the endpoint \(1/4\) is included because \(1/4\in A_4\); including or excluding that endpoint does not alter the interval’s Lebesgue measure.
Worked Example: Moving Spikes with Shrinking Supports
On \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure on the infinite-measure space \(\mathbb{R}\), define \(f_n=n\chi_{[n,n+1/n]}\). Classify convergence to the zero function.
Fix \(x\in\mathbb{R}\). Whenever \(n>x\), the inequality \(x<n\) shows that \(x\notin[n,n+1/n]\); hence \(f_n(x)=0\) for all sufficiently large \(n\). This proves pointwise convergence to zero everywhere, and therefore also convergence almost everywhere. For a fixed \(\varepsilon>0\), if \(n\le\varepsilon\), the value \(n\) does not strictly exceed \(\varepsilon\), so the error set is empty. If \(n>\varepsilon\), the error set is exactly the spike interval:
In the second case its measure is \((n+1/n)-n=1/n\). For each fixed \(\varepsilon\), all sufficiently large \(n\) satisfy \(n>\varepsilon\), and \(1/n\to0\); thus \(f_n\to0\) in measure. Check: for \(\varepsilon=1/2\) and \(n=4\), the spike has height \(4>1/2\), its error set is \([4,4+1/4]=[4,17/4]\), and its measure is \(17/4-4=1/4\), agreeing with \(1/n=1/4\). The conclusion uses the shrinking length of the support, not finiteness of the ambient measure space.
Worked Example: Almost-Everywhere but Not Pointwise Convergence
On \(([0,1],\mathcal{L},m)\), define \(f_n(x)=0\) for \(x\ne0\), and set \(f_n(0)=1\) when \(n\) is even and \(f_n(0)=0\) when \(n\) is odd. Take \(f=0\). At every \(x\in(0,1]\), the sequence is identically zero and converges to \(f(x)\). At \(x=0\), its values alternate between \(1\) and \(0\), so there is no pointwise limit there. Since \(\{0\}\) has Lebesgue measure zero, the sequence converges to \(f\) almost everywhere.
For \(0<\varepsilon<1\), the error set is \(\{0\}\) for even \(n\) and empty for odd \(n\); both have measure zero. For \(\varepsilon\ge1\), it is empty for every \(n\), since the error is at most \(1\) and the threshold comparison is strict. Thus the sequence also converges to zero in measure. Check: at \(\varepsilon=1/2\), the error-set measures for \(n=1,2,3,4\) are \(0,0,0,0\), respectively, so they indeed tend to zero. The failure of pointwise convergence occurs at one point, but that does not prevent either of the other two modes.
Worked Example: Oscillating Indicators with No In-Measure Limit
On the finite measure space \(([0,1],\mathcal{L},m)\), put \(A=[0,1/2]\), \(B=(1/2,1]\), and let \(f_n=\chi_A\) for even \(n\), \(f_n=\chi_B\) for odd \(n\). The sets \(A\) and \(B\) are disjoint and cover \([0,1]\). Therefore at every \(x\), the values \(f_n(x)\) alternate between \(0\) and \(1\). There is no pointwise limit anywhere, so there is no almost-everywhere convergence either.
To rule out convergence in measure, first note a necessary condition. If a sequence of measurable real-valued functions \(f_n\) converges in measure to a measurable real-valued \(f\) on any measure space \((X,\mathcal{M},\mu)\), then for each \(\eta>0\),
Proof. If both \(|f_n(x)-f(x)|\le\eta/2\) and \(|f_m(x)-f(x)|\le\eta/2\), the triangle inequality gives \(|f_n(x)-f_m(x)|\le\eta\). Hence any point where \(|f_n-f_m|>\eta\) must belong to at least one of the two error sets on the right. Taking measures and using subadditivity proves the inequality. If \(f_n\to f\) in measure, then for a fixed \(\eta\), each term on the right tends to zero as its index tends to infinity. Given any \(\delta>0\), choose \(N\) so both measures are less than \(\delta/2\) whenever their indices are at least \(N\). Their sum is then less than \(\delta\), which is the Cauchy condition in measure. \(\square\)
In the oscillating example, for every even \(n\) and odd \(m\), \(|f_n-f_m|=1\) at every point of \([0,1]\). With \(\eta=1/2\), the set \(\{|f_n-f_m|>\eta\}\) is all of \([0,1]\), of measure \(1\), no matter how large \(n\) and \(m\) are. This contradicts the necessary Cauchy condition, so the sequence cannot converge in measure. Check: \(m([0,1])=1-0=1\), and the computed distance is \(1>1/2\) throughout, so the strict-threshold error set really is the whole interval.
Reading the Calculations Correctly
The examples separate several claims that are easy to conflate. Pointwise convergence implies almost-everywhere convergence, because one may choose the exceptional set to be empty. Almost-everywhere convergence does not require convergence at every point, as the alternating values at \(0\) demonstrate. Convergence in measure concerns the measures of error sets at each fixed positive threshold; it does not require the functions to settle down at every point. In the shrinking-interval example, the limit itself is nonzero at \(0\), and the error away from \(0\) occupies a short interval. In the moving-spike example, the supports move away from every fixed point while their measures shrink. In the oscillating example, disagreement between alternating terms occupies the whole space and prevents the sequence from even being Cauchy in measure.
Common mistake: overlooking where the limit differs from the sequence. For the shrinking intervals, one might incorrectly say that the error set is empty because \(f_n=f\) at \(0\). That checks only one point. On every point of \((0,1/n]\), \(f_n=1\) while \(f=\chi_{\{0\}}=0\), so the error is \(1\). For \(0<\varepsilon<1\), the correct error set is \((0,1/n]\), of measure \(1/n\). A second common error is to replace \(>\varepsilon\) by \(\ge\varepsilon\): when an error equals the threshold, it is not in the error set used to test convergence in measure. In particular, an indicator error of \(1\) gives an empty error set at \(\varepsilon=1\), but the symmetric difference is the error set for every threshold strictly between \(0\) and \(1\).
In practice, first identify the candidate limit and establish pointwise behavior directly. For indicators, compare the sets through their symmetric difference. For other sequences, write \(\{|f_n-f|>\varepsilon\}\) explicitly and compute its measure before drawing a conclusion. If convergence in measure seems plausible but no candidate limit is apparent, comparing pairs of terms can expose a failure of the necessary Cauchy condition.
Check Your Understanding
For each question, state the measure space and use the strict error-set definition when calculating convergence in measure.
- On \([0,1]\) with Lebesgue measure, let \(g_n=\chi_{[1/2,\,1/2+1/(2n)]}\). Find its pointwise limit and calculate the error-set measure for \(0<\varepsilon<1\).
- On \(\mathbb{R}\) with Lebesgue measure, let \(h_n=\chi_{[n,n+2]}\). Does \(h_n\) converge pointwise everywhere? Does it converge in measure to zero?
- For the oscillating indicators on \([0,1]\), calculate \(\{|f_n-f_m|>1/2\}\) when \(n\) and \(m\) have the same parity and when they have opposite parity.
- Explain why a sequence that converges pointwise everywhere must converge almost everywhere, but the converse can fail.
- For measurable sets \(A_n,A\), state the exact relationship between the error set for their indicators and \(A_n\,\triangle\,A\) when \(0<\varepsilon<1\).