Height, Width, and Mass
A spike function is large on a small set and zero elsewhere. Its height may increase while the measure of its support decreases, so neither pointwise convergence nor convergence in measure alone determines what happens to its integral. The useful calculation is the product of height and support measure: this is the spike’s mass. The product can remain constant even while the support becomes invisible at every fixed point.
Throughout, \(m\) denotes Lebesgue measure on \(\mathbb{R}\), and the measure space \((\mathbb{R},\mathcal{L},m)\) is sigma-finite and complete, but not finite. All functions in the examples are real-valued and measurable. Pointwise convergence means convergence at every \(x\); convergence \(m\)-a.e. allows failure on a set of measure zero. A sequence \(f_n\) converges to \(f\) in measure if, for every \(\varepsilon>0\), \(m(\{|f_n-f|>\varepsilon\})\to0\). The inequality in this tail set is strict.
For a sequence of spikes \(f_n=a_n\chi_{E_n}\), the integral is therefore controlled by \(a_nm(E_n)\). Pointwise behavior instead depends on which points belong to infinitely many supports. Convergence in measure depends on the measures of the supports at each fixed height threshold. These are three distinct quantities to track.
A Criterion for Spike Sequences
Proof. By definition, \(\limsup E_n=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}E_n\) is the set of points belonging to infinitely many \(E_n\). If \(x\notin\limsup E_n\), then there is an \(N\) such that \(x\notin E_n\) for all \(n\ge N\); hence \(f_n(x)=0\) for all such \(n\), and \(f_n(x)\to0\). If \(x\in\limsup E_n\), then \(x\in E_n\) for infinitely many indices. Along those indices \(f_n(x)=a_n\), and \(a_n\to\infty\), so \(f_n(x)\) cannot converge to zero. Thus the set where pointwise convergence to zero fails is exactly \(\limsup E_n\), proving the first assertion, including the almost-everywhere equivalence. For the tail-set formula, if \(a_n\le\varepsilon\), then \(f_n(x)\le\varepsilon\) for every \(x\), so the strict inequality \(|f_n(x)|>\varepsilon\) never holds. If \(a_n>\varepsilon\), that strict inequality holds exactly when \(x\in E_n\), so the tail set is \(E_n\). The final equivalence is precisely the definition of convergence in measure applied to these computed tail sets. \(\square\)
When \(\mu(E_n)<\infty\), the integral of each spike is \(\int f_n\,d\mu=a_n\mu(E_n)\), by the definition of the integral of a nonnegative simple function. Neither condition in the proposition fixes this product: supports may shrink enough for the product to vanish, or at just the right rate for it to stay positive.
Worked Example: A Spike at the Origin with Constant Mass
On \((\mathbb{R},\mathcal{L},m)\), let \(f_n=n\chi_{(0,1/n)}\). Find its pointwise limit, test convergence in measure, and compute its integral.
At \(x=0\), the indicator is zero for every \(n\), because the interval is open. If \(x<0\), it is also always zero. If \(x>0\), choose an integer \(N>1/x\). For \(n\ge N\), \(1/n\le1/N<x\), so \(x\notin(0,1/n)\). Thus \(f_n(x)=0\) eventually at every \(x\in\mathbb{R}\), and the pointwise limit is the zero function everywhere.
Fix \(\varepsilon>0\). If \(n\le\varepsilon\), then \(f_n(x)\le\varepsilon\) for every \(x\), so the strict tail set \(\{|f_n|>\varepsilon\}\) is empty. If \(n>\varepsilon\), that tail set is exactly \((0,1/n)\). Consequently, once \(n>\varepsilon\),
Therefore \(f_n\to0\) in measure. Its integral, however, is
Check: at \(n=4\), the height is \(4\), the interval has measure \(1/4\), and the integral is \(4\cdot(1/4)=1\). Direct integration also gives \(\int_0^{1/4}4\,dx=[4x]_0^{1/4}=1\). The limit function has integral zero, so the integrals do not converge to the integral of the pointwise limit.
Supports That Move Away
A support need not shrink toward a fixed point to produce pointwise convergence. It can move out along the real line. If its measure shrinks, the sequence can still converge in measure; if the support keeps a fixed measure, pointwise convergence need not imply convergence in measure on this infinite-measure space.
Worked Example: A Narrow Spike Escaping to Infinity
Let \(g_n=n\chi_{(n,n+1/n)}\) on \((\mathbb{R},\mathcal{L},m)\). Determine the pointwise limit, convergence in measure, and integrals.
For each fixed \(x\in\mathbb{R}\), membership in \((n,n+1/n)\) requires \(n<x\). There are only finitely many positive integers \(n<x\); for all larger \(n\), \(g_n(x)=0\). Hence \(g_n(x)\to0\) at every \(x\). For any fixed \(\varepsilon>0\), when \(n>\varepsilon\) the strict tail set \(\{|g_n|>\varepsilon\}\) is \((n,n+1/n)\), whose measure is \(1/n\). Thus \(g_n\to0\) in measure.
The integral is
Check: for \(n=3\), the support is \((3,10/3)\), with length \(10/3-3=1/3\); height \(3\) times length \(1/3\) gives integral \(1\). Direct integration over that interval gives \([3x]_3^{10/3}=10-9=1\). The mass stays positive even though every fixed point eventually lies outside the support: it has escaped to larger locations.
Worked Example: Pointwise Convergence Without Convergence in Measure
Let \(h_n=\chi_{(n,n+1)}\) on \((\mathbb{R},\mathcal{L},m)\). Show why its pointwise convergence to zero does not give convergence in measure on all of \(\mathbb{R}\).
For any fixed \(x\), the condition \(x\in(n,n+1)\) implies \(n<x\), which can hold for only finitely many positive integers \(n\). Therefore \(h_n(x)\) is eventually zero, and \(h_n\to0\) pointwise everywhere. Choose the fixed threshold \(\varepsilon=1/2\). Since the function takes only the values \(0\) and \(1\), and \(1>1/2\), the strict tail set is \((n,n+1)\) for every \(n\). Thus
for every \(n\). The tail measures do not tend to zero, so \(h_n\) does not converge to zero in measure. Its integral is also constant: \(\int h_n\,dm=m((n,n+1))=1\). Check: at \(n=2\), the support \((2,3)\) has measure \(3-2=1\), and its indicator integrates to \(1\). This example differs from a narrowing spike: the support moves, but its measure does not decrease.
When the Spike Mass Vanishes
The same height-width calculation also gives convergence of integrals when the support shrinks faster than the height grows. This is a separate conclusion from convergence in measure: the integral tests the product of height and width, whereas the tail-set test examines the width once the height exceeds the chosen threshold.
Worked Example: Increasing Height with Vanishing Mass
On \((\mathbb{R},\mathcal{L},m)\), set \(q_n=\sqrt{n}\chi_{(0,1/n)}\). Find the pointwise limit, convergence in measure, and integrals.
As in the first example, at each \(x>0\), the inequality \(x<1/n\) eventually fails; at \(x\le0\), it always fails. Therefore \(q_n\to0\) everywhere. For a fixed \(\varepsilon>0\), whenever \(\sqrt n>\varepsilon\), the strict tail set \(\{|q_n|>\varepsilon\}\) is \((0,1/n)\), and its measure is \(1/n\to0\). Hence \(q_n\to0\) in measure.
Each integral is
Check: at \(n=9\), the height is \(\sqrt9=3\), the support length is \(1/9\), and the integral is \(3\cdot(1/9)=1/3\), matching \(1/\sqrt9=1/3\). Direct integration gives \([3x]_0^{1/9}=1/3\). Here the mass vanishes, unlike the constant-mass spikes.
Common Pitfall: Confusing Height with Mass
A common incorrect calculation is to say that the first spike has integral \(n\) because its height is \(n\). That omits the support measure. The correct calculation is
Another mistake is to infer convergence in measure just from pointwise convergence on an infinite-measure space. In the moving-interval example, pointwise convergence holds, but at the same threshold \(1/2\) every tail set has measure \(1\). A reliable check is to choose a fixed \(\varepsilon>0\), compute \(m(\{|f_n|>\varepsilon\})\) with the strict inequality, and then take its limit.
These spike calculations isolate why pointwise behavior, convergence in measure, and integral behavior must be checked separately. A shrinking support can make the tail sets small; an escaping support can make each fixed point eventually unaffected; and the height-times-measure product can still remain constant. Conversely, when that product tends to zero, the integrals of these nonnegative spikes tend to the integral of their zero pointwise limit.
Check Your Understanding
For each sequence, calculate the support measure and the integral before deciding which convergence statements hold.
- For \(r_n=2n\chi_{(0,1/(2n))}\) on \(\mathbb{R}\), determine the pointwise limit, convergence in measure, and integral.
- For \(s_n=3\chi_{(n,n+2)}\), test convergence in measure to zero using the fixed strict threshold \(\varepsilon=1\).
- For \(t_n=n^2\chi_{(0,1/n^3)}\), compute \(\int t_n\,dm\) and decide whether it tends to zero.
- State the set whose nullity characterizes pointwise convergence to zero for \(a_n\chi_{E_n}\) when \(a_n\to\infty\).
- Explain why pointwise convergence of \(\chi_{(n,n+1)}\) on \(\mathbb{R}\) does not imply convergence in measure there.