Decreasing Limits Need a Finite Starting Size
For an increasing sequence of nonnegative measurable functions, the Monotone Convergence Theorem identifies the integral of the limit with the limit of the integrals, even when those values are infinite. Decreasing sequences behave differently: without a finite bound on the first integral, their integrals need not approach the integral of the pointwise limit. The basic example is the sequence of indicators of the tails \([n,\infty)\) on the real line. Each tail has infinite Lebesgue measure, but their intersection is empty.
We use \((\mathbb{R},\mathcal{L},m)\), Lebesgue measure on the Lebesgue sigma-algebra; this measure space is complete, sigma-finite, and not finite. For a measurable set \(E\), its indicator \(\chi_E\) is the real-valued function equal to \(1\) on \(E\) and \(0\) outside \(E\). Its Lebesgue integral is \(\int\chi_E\,dm=m(E)\), with \(+\infty\) allowed. More generally, statements about functions below concern measurable functions with values in \([0,\infty]\), and limits are pointwise unless explicitly stated otherwise.
Continuity from Above for Sets
Proof. Set \(E=\bigcap_{n=1}^{\infty}E_n\) and \(F_n=E_1\setminus E_n\). The sets \(F_n\) are measurable and increasing, and their union is \(E_1\setminus E\). To see the latter equality, a point of \(E_1\setminus E\) fails to belong to some \(E_k\), so it belongs to \(E_1\setminus E_k=F_k\); conversely, no point of any \(F_k\) belongs to \(E\). By continuity from below for measures, \(\mu(F_n)\) increases to \(\mu(E_1\setminus E)\). Since \(\mu(E_1)<\infty\), all the following subtractions are finite: $$ \mu(E_n)=\mu(E_1)-\mu(F_n) \quad\hbox{and}\quad \mu(E)=\mu(E_1)-\mu(E_1\setminus E). $$ Taking the limit in the first equality gives the second. This proves the claimed identity, including the case \(\mu(E_1)=0\). \(\square\)
The finite hypothesis is used in the subtraction: it prevents an undefined expression of the form \(\infty-\infty\). More importantly, without it the conclusion itself can fail.
Worked Example: The Tails [n, infinity)
For each positive integer \(n\), let \(E_n=[n,\infty)\subset\mathbb{R}\). These are Borel, hence Lebesgue measurable, and \(E_{n+1}\subseteq E_n\). Their intersection is empty: for any fixed real \(x\), choose an integer \(n>x\); then \(x\notin[n,\infty)\).
Each tail has infinite Lebesgue measure. Indeed, for any integer \(N>n\), the interval \([n,N]\) is contained in \(E_n\) and has measure \(N-n\). These lengths have no finite upper bound as \(N\) increases, so \(m(E_n)=\infty\). Therefore $$ \int_{\mathbb{R}}\chi_{E_n}\,dm=m(E_n)=\infty \quad\hbox{for every }n, \qquad \int_{\mathbb{R}}\chi_{\cap_nE_n}\,dm =\int_{\mathbb{R}}0\,dm=0. $$ The functions \(\chi_{E_n}\) decrease pointwise to zero, while their integrals have limit \(+\infty\), not \(0\). Check: at a fixed \(x\), the indicator is eventually zero because every integer \(n>x\) has \(x\notin E_n\); this confirms the pointwise limit used in the calculation.
This does not contradict the Monotone Convergence Theorem: the indicators decrease, rather than increase. It also shows why the statement “decreasing sets with empty intersection have measures tending to zero” is not valid without a finite-measure assumption.
What the Finite Hypothesis Gives for Integrals
Proof. Since the sequence decreases pointwise, \(f(x)=\inf_n f_n(x)\), so \(f\) is measurable. Also \(0\le f\le f_1\), which gives \(\int f\,d\mu\le\int f_1\,d\mu<\infty\). To apply the Monotone Convergence Theorem to differences without encountering \(\infty-\infty\), let \(N=\{x:f_1(x)=\infty\}\). This set is measurable and has measure zero: for every positive integer \(k\), \(k\chi_N\le f_1\), so \(k\mu(N)\le\int f_1\,d\mu<\infty\); letting \(k\) increase forces \(\mu(N)=0\).
Define \(\widetilde f_n=f_n\) on \(X\setminus N\) and \(\widetilde f_n=0\) on \(N\), and define \(\widetilde f\) in the same way from \(f\). These functions are measurable, finite-valued, and agree with the original functions outside a measurable null set, so their integrals are unchanged. Now \(g_n=\widetilde f_1-\widetilde f_n\) is a nonnegative measurable function. Because \(\widetilde f_n\) decreases, \(g_n\) increases, and its pointwise limit is \(g=\widetilde f_1-\widetilde f\). By the Monotone Convergence Theorem, \(\int g_n\,d\mu\to\int g\,d\mu\). The additivity of the nonnegative Lebesgue integral gives $$ \int\widetilde f_n\,d\mu =\int\widetilde f_1\,d\mu-\int g_n\,d\mu \longrightarrow \int\widetilde f_1\,d\mu-\int g\,d\mu =\int\widetilde f\,d\mu. $$ All subtractions here are finite because \(0\le g_n\le\widetilde f_1\) and \(\int\widetilde f_1\,d\mu<\infty\). The original and modified functions have the same integrals, which proves the result. \(\square\)
The pointwise monotonicity assumption in this statement avoids any measurability issue for the limit: a pointwise limit of measurable functions is measurable. If monotonicity and convergence are assumed only almost everywhere on an incomplete measure space, the almost-everywhere limit need not be measurable on the exceptional null set. In that formulation one must also require the limit to be measurable, or assume completeness.
Worked Example: Finite Lebesgue Integrals on a Tail
On \(X=[0,\infty)\) with the restricted Lebesgue measure, define \(f_n(x)=e^{-x}\chi_{[n,\infty)}(x)\) for positive integers \(n\). These functions are measurable, nonnegative, and decrease pointwise, since the tails shrink. The first function is integrable: $$ \int_0^\infty f_1(x)\,dx =\int_1^\infty e^{-x}\,dx =\left[-e^{-x}\right]_{1}^{\infty} =e^{-1}<\infty. $$ For each \(n\), the same antiderivative gives $$ \int_0^\infty f_n(x)\,dx =\int_n^\infty e^{-x}\,dx =\left[-e^{-x}\right]_{n}^{\infty} =e^{-n}. $$ At each fixed \(x\ge0\), \(f_n(x)=0\) whenever \(n>x\), so \(f_n(x)\to0\). The computed integrals satisfy \(e^{-n}\to0\), agreeing with the integral of the limit. Check: substituting \(n=1\) gives the first integral \(e^{-1}\), as computed directly, and the sequence \(e^{-n}\) tends to zero because \(e^{-n}=(e^{-1})^n\) with \(0<e^{-1}<1\).
The Pitfall: Treating Infinite Measure Like Finite Measure
A common incorrect calculation is: “\(E_n\downarrow\varnothing\), therefore \(m(E_n)\to m(\varnothing)=0\).” For \(E_n=[n,\infty)\), this would assert \(\infty\to0\). The correction is to check the hypothesis before using continuity from above: \(m(E_1)=m([1,\infty))=\infty\), so the theorem does not apply. In fact every term of the sequence of measures is \(+\infty\), and its limit in the extended real numbers is \(+\infty\).
Likewise, pointwise convergence of nonnegative functions alone does not authorize passing a limit through an integral. In the tail example, \(\chi_{[n,\infty)}\to0\) pointwise but the integrals remain infinite. The finite-integral theorem applies when the first function has finite integral and the sequence decreases everywhere; it does not claim that every decreasing sequence has convergent integrals. The distinction is the same for indicators: the finite-measure condition on the first set is what makes continuity from above valid.
Worked Example: A Decreasing Sequence with a Nonempty Intersection
Let \(E_n=[0,1+1/n]\) in \((\mathbb{R},\mathcal{L},m)\), for positive integers \(n\). Since \(1+1/(n+1)<1+1/n\), these measurable intervals decrease. A point belongs to every \(E_n\) exactly when it lies between \(0\) and \(1\): points larger than \(1\) eventually exceed \(1+1/n\), while every point in \([0,1]\) belongs to all the intervals. Hence \(\bigcap_nE_n=[0,1]\). The measures and indicator integrals are $$ m(E_n)=1+\frac1n,\qquad \int_{\mathbb{R}}\chi_{E_n}\,dm=1+\frac1n, \qquad m\left(\bigcap_nE_n\right)=\int_{\mathbb{R}}\chi_{[0,1]}\,dm=1. $$ Thus the integrals decrease to \(1\), the integral of the limiting indicator. The finite hypothesis holds because \(m(E_1)=2\). Check: at \(n=1\), the interval is \([0,2]\) and has measure \(2\); the formula \(1+1/n\) gives \(2\) as well.
These results are useful whenever a decreasing sequence represents successive restrictions, truncations, or tail events. Before taking the limit of their measures or integrals, identify a finite first measure or an integrable first function. When that control is absent, the tail indicators on the real line show that even a pointwise limit equal to zero can coexist with integrals that never become finite.
Check Your Understanding
For each question, identify whether the finite-measure or finite-integral hypothesis is available before drawing a conclusion.
- For \(E_n=[n,\infty)\), compute \(\bigcap_nE_n\), \(m(E_n)\), and the limit of the indicator integrals.
- State the finite-measure hypothesis in continuity from above and identify exactly where it is used in the proof.
- For \(E_n=[0,1/n]\), find the intersection and verify continuity from above by computing every measure.
- Why is \(f_n=e^{-x}\chi_{[n,\infty)}\) covered by the decreasing-integral theorem on \([0,\infty)\), while \(\chi_{[n,\infty)}\) on \(\mathbb{R}\) is not?
- If \(f_n\) decreases pointwise and \(\int f_1\,d\mu<\infty\), what does the theorem imply about the measurability and integrability of \(f=\lim_nf_n\)?