From Lower Simple Estimates to Limits
The nonnegative integral was defined as the supremum of the integrals of measurable simple functions lying below the integrand. The Increasing Simple Approximation theorem established earlier gives one way to build such lower estimates: a nonnegative measurable function can be approached pointwise from below by an increasing sequence of simple functions. The question now is how the integrals behave when the functions themselves increase.
The answer is the Monotone Convergence Theorem. It says that for an increasing sequence of nonnegative measurable functions, integrating the pointwise limit gives the same result as taking the limit of the integrals. The theorem allows the value \(+\infty\); neither the limiting function nor its integral must be finite. Its proof follows in the next tutorial. Here we state the result, examine its conditions, and work through consequences and examples.
The condition \(f_n(x)\leq f_{n+1}(x)\) is pointwise and must hold at every \(x\). Since the values are nonnegative, the pointwise limit exists in \([0,\infty]\), and the Pointwise Limits of Measurable Functions theorem ensures that \(f\) is measurable. The theorem therefore compares integrals of measurable functions, including the limit.
There is already a preliminary consequence of the hypothesis. By the Monotonicity of the Lebesgue Integral theorem from earlier in the course, $$ \int_X f_n\,d\mu\leq\int_X f_{n+1}\,d\mu $$ for each \(n\). Thus the integrals form a nondecreasing sequence in \([0,\infty]\), so their limit is well-defined. The Monotone Convergence Theorem identifies that limit with the integral of the pointwise limit. It is the equality between these two quantities—not merely the fact that the integrals increase—that makes the theorem powerful.
A Set-Theoretic Special Case
Taking indicator functions turns pointwise increase into inclusion of measurable sets. This gives a useful special case of the theorem, and also connects it to Continuity from Below, established earlier in the course.
Proof. Fix \(x\in X\). If \(x\in A\), then \(x\in A_N\) for some \(N\). Because the sets increase, \(x\in A_n\) for every \(n\geq N\), so \(\mathbf{1}_{A_n}(x)\) is eventually \(1\). If \(x\notin A\), then \(x\notin A_n\) for every \(n\), so \(\mathbf{1}_{A_n}(x)=0\) for every \(n\). In either case, $$ \lim_{n\to\infty}\mathbf{1}_{A_n}(x)=\mathbf{1}_A(x). $$ The indicators are measurable and increase pointwise. Applying the Monotone Convergence Theorem gives the asserted convergence of their integrals. By the Integral of an Indicator theorem, this equality is also \(\mu(A_n)\to\mu(A)\), which is precisely Continuity from Below. \(\square\)
This special case makes the structure of the general theorem visible: regions on which the approximating functions are positive can expand, and the integral records the accumulated measure-weighted contribution. The full theorem handles changing function values as well as changing regions.
Worked Example: Indicators of Expanding Intervals
On \(\mathbb{R}\) with Lebesgue measure, let \(A_n=[-n,n]\) and \(f_n=\mathbf{1}_{A_n}\). The intervals are nested, and their union is \(\mathbb{R}\). Hence \(f_n(x)\) increases to \(1\) for every real \(x\). For each \(n\), $$ \int_{\mathbb{R}}f_n\,d\lambda =\lambda([-n,n]) =2n. $$ The integrals tend to \(+\infty\), and the limit function is \(\mathbf{1}_{\mathbb{R}}\), whose integral is \(\lambda(\mathbb{R})=\infty\). Thus both sides of the Monotone Convergence Theorem equal \(+\infty\). This example also shows why the theorem must allow infinite integrals.
Worked Example: Increasing Indicators with a Finite Limit
On \([0,1]\) with Lebesgue measure, let \(A_n=[0,1-1/n]\) for positive integers \(n\). In particular, \(A_1=[0,0]=\{0\}\), and the sets increase. Their union is \([0,1)\): every \(x<1\) in \([0,1]\) belongs to \(A_n\) once \(1/n\leq 1-x\), while \(1\) belongs to none of the sets. Therefore \(\mathbf{1}_{A_n}\) increases pointwise to \(\mathbf{1}_{[0,1)}\).
For every \(n\), the interval has measure \(1-1/n\), including \(n=1\), when this is zero. Consequently, $$ \int_{[0,1]}\mathbf{1}_{A_n}\,d\lambda =\lambda(A_n) =1-\frac{1}{n} \longrightarrow 1. $$ The limit function has integral \(\lambda([0,1))=1\), in agreement with the theorem. The endpoint \(1\) is not in the union, but its omission does not change the measure.
Increasing Function Values, Not Just Increasing Domains
The indicator case changes only the sets on which the function equals \(1\). The Monotone Convergence Theorem also applies when the values rise at points. A particularly useful setting is an increasing sequence of simple functions: their integrals can be calculated directly, while the theorem identifies the integral of their pointwise limit.
Worked Example: Geometric Values on Counting Measure
Take \(X=\mathbb{N}\) with the power-set sigma-algebra and counting measure. For each \(n\), define $$ f_n(k)= \begin{cases} 2^{-k},&1\leq k\leq n,\\ 0,&k>n. \end{cases} $$ Each \(f_n\) is a nonnegative measurable simple function. For every fixed \(k\), its value is zero when \(n<k\) and is \(2^{-k}\) when \(n\geq k\); hence \(f_n(k)\leq f_{n+1}(k)\). Its pointwise limit is \(f(k)=2^{-k}\).
Under counting measure, the integral of a nonnegative simple function is the sum of its values, so $$ \int_{\mathbb{N}} f_n\,d\mu =\sum_{k=1}^{n}2^{-k} =1-2^{-n}. $$ The finite-sum identity follows by multiplying the sum by \(2\) and subtracting the original sum: $$ 2\sum_{k=1}^{n}2^{-k}-\sum_{k=1}^{n}2^{-k} =1-2^{-n}. $$ As \(n\to\infty\), these integrals tend to \(1\). The Monotone Convergence Theorem therefore gives \(\int_{\mathbb{N}}f\,d\mu=1\). The finite-stage functions capture more and more of the limiting function’s total integral.
Using Truncations for an Extended-Valued Function
The Increasing Truncation Approximation theorem from earlier in the course says that for an extended nonnegative measurable function \(f\), the functions \(f\wedge n=\min(f,n)\) increase pointwise to \(f\). The Monotone Convergence Theorem consequently gives a useful formula: $$ \int_X f\,d\mu=\lim_{n\to\infty}\int_X(f\wedge n)\,d\mu. $$ This includes functions that equal \(+\infty\) on part of the space. It does not say that every truncation has a finite integral; the measure space may have infinite measure, or the truncation may be positive on a set of infinite measure.
Worked Example: An Infinite Value on a Null Set
Let \(X=[0,1]\) with Lebesgue measure, and define \(f(0)=+\infty\) and \(f(x)=1\) for \(0<x\leq1\). For each positive integer \(n\), let \(f_n=f\wedge n\). Since \(n\geq1\), $$ f_n= n\mathbf{1}_{\{0\}}+\mathbf{1}_{(0,1]}. $$ These are measurable simple functions. At \(x=0\), their values are \(1,2,3,\ldots\), so they increase to \(+\infty\); at every \(x\in(0,1]\), their value is constantly \(1\). Thus \(f_n\) increases pointwise to \(f\).
The singleton \(\{0\}\) has Lebesgue measure zero, while \((0,1]\) has measure \(1\). The simple-function integral formula gives, for every \(n\), $$ \int_{[0,1]}f_n\,d\lambda =n\lambda(\{0\})+\lambda((0,1]) =n\cdot0+1 =1. $$ The limit of these integrals is \(1\). By the Monotone Convergence Theorem, \(\int_{[0,1]}f\,d\lambda=1\), even though \(f(0)=+\infty\). A value at a single null point does not contribute to the integral.
Why the Increasing Hypothesis Matters
The theorem requires the functions to increase pointwise. Pointwise convergence alone does not guarantee convergence of the integrals. Nor is it enough for the integrals to increase: the relationship between the functions at each point is a distinct hypothesis. The following sequence illustrates a common mistake in checking that condition.
Worked Example: A Sequence That Decreases Instead
On \([0,1]\), define \(g_n=\mathbf{1}_{(0,1/n)}\). The intervals \((0,1/n)\) shrink, so the sequence of indicators is nonincreasing, not nondecreasing. In particular, for any fixed \(x\) with \(0<x<1\), \(g_1(x)=1\), because \(x\in(0,1)\). Once \(n\geq 1/x\), we have \(1/n\leq x\), so \(x\notin(0,1/n)\) and \(g_n(x)=0\). Thus at such a point the sequence is eventually zero after having been positive.
At the endpoints, \(g_n(0)=g_n(1)=0\) for every \(n\), since neither endpoint belongs to \((0,1/n)\). At every point of \([0,1]\), the pointwise limit is zero. Also, $$ \int_{[0,1]}g_n\,d\lambda =\lambda((0,1/n)) =\frac{1}{n} \longrightarrow 0. $$ This particular sequence happens to have integrals converging to the integral of its limit, but it does not satisfy the increasing hypothesis. Its convergence cannot be justified by the Monotone Convergence Theorem. Checking a point in \((0,1)\), rather than only the endpoints, is essential to seeing why the sequence is not increasing.
A useful distinction is that monotone convergence is a sufficient condition, not a necessary one: some sequences outside its hypotheses may still have convergent integrals. The theorem guarantees the desired equality when the functions increase pointwise; it makes no assertion about every other kind of pointwise convergence.
Check Your Understanding
Use the theorem’s hypotheses and examples to check your understanding.
- What pointwise condition must hold between each \(f_n\) and \(f_{n+1}\) in the Monotone Convergence Theorem?
- Why is the sequence of integrals nondecreasing when the functions increase pointwise?
- How does the Monotone Convergence Theorem for increasing indicators relate to Continuity from Below?
- For the geometric-valued functions on \(\mathbb{N}\), what is the pointwise limit and what are the finite-stage integrals?
- In the truncation example, why does the value \(f(0)=+\infty\) not make the integral infinite?
- For \(g_n=\mathbf{1}_{(0,1/n)}\), what happens at a fixed point \(0<x<1\), and why does this prevent applying the theorem?