Infinite Values in Monotone Convergence

When the Limit or Integral Is Infinite

The Monotone Convergence Theorem is an equality in the extended nonnegative real numbers: either side may be \(+\infty\). Thus an increasing sequence can converge to a function whose integral is infinite, even when every function in the sequence has a finite integral. There is a different possibility: the limit function may equal \(+\infty\) on a set of positive measure, which by itself forces its integral to be infinite. Keeping these two mechanisms separate is essential.

Throughout, \(([0,1],\mathcal{L},m)\) is the Lebesgue measure space on the unit interval, unless another space is specified. Functions in the examples are measurable and take values in \([0,\infty]\). The nonnegative Lebesgue integral is allowed to have value \(+\infty\). A statement holding almost everywhere means that it may fail on a set of Lebesgue measure zero.

Monotone Convergence Theorem (available result): On any measure space \((X,\mathcal{M},\mu)\), if \(f_n:X\to[0,\infty]\) are measurable and \(f_n\le f_{n+1}\) everywhere, with \(f=\lim_{n\to\infty}f_n\), then \(f\) is measurable and \(\int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu\). Both sides are allowed to be \(+\infty\). This theorem was proved in the earlier material on monotone convergence.

The sequence of integrals in this theorem is nondecreasing, but its terms need not be finite. If they are all finite while their limit is infinite, the equality still applies without change. Nor does the theorem require \(f\) to be finite almost everywhere.

Two Consequences for Infinite Integrals

Proposition: Let \((X,\mathcal{M},\mu)\) be a measure space, and let \(f_n:X\to[0,\infty]\) be measurable with \(f_n\uparrow f\) everywhere. Then \(\int f\,d\mu=+\infty\) if and only if \(\lim_{n\to\infty}\int f_n\,d\mu=+\infty\). In particular, the conclusion includes the case where some \(\int f_n\,d\mu\) is already infinite.

Proof. By the Monotone Convergence Theorem, \(\int f\,d\mu=\lim_n\int f_n\,d\mu\) as an equality in \([0,\infty]\). Therefore the left side is \(+\infty\) exactly when the limit on the right is \(+\infty\). Since the integrals form a nondecreasing sequence, if one term is \(+\infty\), every subsequent term is \(+\infty\) as well. \(\square\)

Proposition: Let \(f:X\to[0,\infty]\) be measurable on a measure space \((X,\mathcal{M},\mu)\), and set \(A=\{x\in X:f(x)=+\infty\}\). If \(\mu(A)>0\), then \(\int f\,d\mu=+\infty\). This implication has no converse: an infinite integral need not come from a positive-measure set where \(f=+\infty\).

Proof. For each positive integer \(k\), the simple function \(k\chi_A\) is measurable and satisfies \(0\le k\chi_A\le f\), including on \(A\), where \(f=+\infty\). Monotonicity of the nonnegative integral gives \(\int f\,d\mu\ge k\mu(A)\). If \(0<\mu(A)<\infty\), these lower bounds grow without bound as \(k\to\infty\). If \(\mu(A)=+\infty\), already \(\int k\chi_A\,d\mu=+\infty\) for every \(k\ge1\). In either case, \(\int f\,d\mu=+\infty\). The examples below show that the converse fails. \(\square\)

These propositions distinguish the size of the set where the function is infinite from the size of its integral. An infinite value at one point of \([0,1]\) can contribute no integral at all, while a function finite almost everywhere can still have infinite integral because its values become large near that point.

Worked Example: A Singular Limit with Infinite Integral

For each positive integer \(n\), define \(f_n:[0,1]\to[0,\infty)\) by \(f_n(0)=n\) and, for \(0<x\le1\), \(f_n(x)=\min(n,1/x)\). Find the pointwise limit and compute every \(\int f_n\,dm\).

For \(x=0\), \(f_n(0)=n\) increases to \(+\infty\). For fixed \(x>0\), the sequence \(\min(n,1/x)\) is nondecreasing and eventually equals \(1/x\). Thus \(f_n\uparrow f\), where \(f(0)=+\infty\) and \(f(x)=1/x\) for \(0<x\le1\). The functions are measurable because they are formed from measurable functions by taking a minimum, with a separately specified value at the measurable singleton \(\{0\}\).

On \(0<x\le1/n\), \(1/x\ge n\), so \(f_n(x)=n\). On \(1/n\le x\le1\), \(1/x\le n\), so \(f_n(x)=1/x\). The formulas agree at \(x=1/n\). The value at the single endpoint \(0\) does not affect the Lebesgue integral. Therefore

\int_0^1 f_n\,dm =\int_0^{1/n}n\,dx+\int_{1/n}^1\frac{1}{x}\,dx =n\cdot\frac1n+\bigl[\ln x\bigr]_{1/n}^1 =1+\ln n.

Each integral is finite, but \(1+\ln n\to+\infty\). By the Monotone Convergence Theorem, \(\int_0^1 f\,dm=+\infty\). The divergence is not caused by the value at \(0\) alone: that singleton has measure zero. Indeed, for every nonnegative integer \(j\), the intervals \((2^{-(j+1)},2^{-j}]\) are pairwise disjoint, and on them \(f(x)=1/x\). The integral over each interval is

\int_{2^{-(j+1)}}^{2^{-j}}\frac{1}{x}\,dx =\ln(2^{-j})-\ln(2^{-(j+1)}) =\ln 2.

Consequently, the \(k\) disjoint intervals with \(j=0,\ldots,k-1\) have total integral \(k\ln2\), which is unbounded as \(k\to\infty\). Check: substituting \(n=2\) in the truncation formula gives \(1+\ln2\); directly, the two pieces have integrals \(2\cdot(1/2)=1\) and \(\int_{1/2}^1 dx/x=\ln2\), the same total.

Worked Example: Infinity on a Set of Positive Measure

On \(([0,1],\mathcal{L},m)\), define \(g_n=n\chi_{[0,1/2]}\). Find its pointwise limit and the limit of its integrals.

Each \(g_n\) is nonnegative and measurable. At every \(x\in[0,1/2]\), \(g_n(x)=n\to+\infty\); at every \(x\in(1/2,1]\), \(g_n(x)=0\). Hence \(g_n\uparrow g\), where \(g=+\infty\) on \([0,1/2]\) and \(g=0\) elsewhere. The set \([0,1/2]\) has measure \(1/2\), so the proposition gives \(\int g\,dm=+\infty\). Directly,

\int_0^1g_n\,dm=n\,m([0,1/2])=n\cdot\frac12=\frac n2 \quad\longrightarrow\quad+\infty.

The Monotone Convergence Theorem confirms that this limit is \(\int g\,dm\). Check: for \(n=2\), the integral is \(2\cdot(1/2)=1\); for \(n=4\), it is \(4\cdot(1/2)=2\). The values increase without bound, as required. The endpoint \(1/2\) is included in the set, but changing membership of that single endpoint would not change its Lebesgue measure or these integrals.

Worked Example: An Infinite Value on a Null Set with Finite Integral

For each positive integer \(n\), define \(h_n(0)=n\) and \(h_n(x)=\min(n,x^{-1/2})\) for \(0<x\le1\). Compute \(\lim_n\int_0^1h_n\,dm\) and identify the pointwise limit.

At \(x=0\), the values increase to \(+\infty\); for \(x>0\), they increase to \(x^{-1/2}\). Thus \(h_n\uparrow h\), where \(h(0)=+\infty\) and \(h(x)=x^{-1/2}\) for \(x>0\). For \(0<x\le n^{-2}\), \(x^{-1/2}\ge n\), so \(h_n(x)=n\); for \(n^{-2}\le x\le1\), \(h_n(x)=x^{-1/2}\). The two formulas agree at \(n^{-2}\). The point \(0\) has measure zero, so

\begin{aligned} \int_0^1h_n\,dm &=\int_0^{n^{-2}}n\,dx+\int_{n^{-2}}^1x^{-1/2}\,dx\\ &=n\cdot n^{-2}+\bigl[2\sqrt{x}\bigr]_{n^{-2}}^1\\ &=\frac1n+2\left(1-\frac1n\right) =2-\frac1n. \end{aligned}

As \(n\to\infty\), these integrals tend to \(2\). Monotone convergence gives \(\int_0^1h\,dm=2\), even though \(h(0)=+\infty\). The value agrees with the improper integral \(\int_0^1x^{-1/2}\,dx=2\); the value at \(0\) does not affect the Lebesgue integral. Check: when \(n=1\), \(h_1=1\) on all of \([0,1]\), including \(0\), so its integral is \(1\), matching \(2-1/1=1\).

A Common Mistake: One Infinite Value Does Not Force Infinite Integral

A tempting but incorrect argument is: “The limit equals \(+\infty\) at \(0\), so its integral is infinite.” For a concrete counterexample, set \(q_n=n\chi_{\{0\}}\) on \(([0,1],\mathcal{L},m)\). Then \(q_n\uparrow q\), where \(q(0)=+\infty\) and \(q(x)=0\) for \(x>0\). Since \(m(\{0\})=0\),

\int_0^1q_n\,dm=n\,m(\{0\})=n\cdot0=0 \quad\text{for every }n, \qquad \int_0^1q\,dm=0.

The correction is to ask how large the set of infinite values is and how the function behaves away from it. If the function is infinite on a set of positive measure, the integral is infinite. If it is infinite only on a null set, its integral may be finite, as \(h\) shows, or infinite, as the limit \(f\) shows. Another incorrect inference is that finite integrals at every stage guarantee a finite limiting integral. The values \(1+\ln n\) in the first example disprove that inference: each is finite, but they are unbounded.

How to Read an Infinite-Value Limit

When applying monotone convergence in a problem with extended values, check the pointwise behavior and the integrals separately. First verify that every function is measurable, nonnegative, and increasing pointwise; then identify the limit, allowing \(+\infty\). Next compute or bound the integrals at finite stages. If those integrals tend to \(+\infty\), the theorem says the limiting integral is \(+\infty\); if their limit is finite, it gives that finite value even if the limit function is infinite at some points.

These conclusions are about the Lebesgue integral, not a claim that the limit function must be bounded or finite everywhere. The examples exhibit all three relevant outcomes: a limit infinite on a positive-measure set has infinite integral; a limit infinite at a null point can have finite integral; and a limit infinite only at a null point can nonetheless have infinite integral because of its behavior on every neighborhood of that point.

Check Your Understanding

For each question, specify whether the relevant limit or integral is finite or infinite, and justify your conclusion.

On \([0,1]\), let \(u_n=n\chi_{[1/3,1]}\). Find the pointwise limit and compute \(\int_0^1u_n\,dm\) for each \(n\).
Let \(v_n(0)=n\) and \(v_n(x)=\min(n,1/\sqrt{x})\) for \(x>0\). Where is the limit infinite, and what is the limit of the integrals?
Give an increasing sequence of nonnegative measurable functions whose limit is infinite at one point but whose integrals are all zero.
Can a sequence of finite integrals have limit \(+\infty\) under monotone convergence? Identify an example above and give its integral formula.
If a nonnegative measurable function is infinite on a set of positive measure, explain why its integral must be infinite.

Infinite Values in Monotone Convergence

What You'll Learn