A Checklist Before Passing a Limit Through an Integral
Pointwise convergence alone does not justify the identity \(\lim_n\int f_n\,d\mu=\int\lim_n f_n\,d\mu\). Before using such an identity, check that the functions and proposed limit are measurable, that the stated mode of convergence actually holds, and that a theorem applicable to the sequence supplies the conclusion. In particular, a bound on the values of \(f_n\) is not enough: for the Dominated Convergence Theorem, the common bound must itself be integrable.
Unless an example specifies otherwise, let \((X,\mathcal{M},\mu)\) be a measure space, not assumed finite, sigma-finite, or complete. Functions in the discussion of the Dominated Convergence Theorem are real-valued and measurable. A function \(g\) is integrable when it is measurable and \(\int |g|\,d\mu<\infty\); for a nonnegative measurable \(g\), this means \(\int g\,d\mu<\infty\). An assertion holding \(\mu\)-almost everywhere (\(\mu\)-a.e.) may fail on a measurable set of measure zero. These qualifications matter: on an incomplete measure space, an almost-everywhere limit need not be measurable if its values on a null set are chosen arbitrarily.
The Dominated Convergence Theorem applies on arbitrary measure spaces; finite measure and sigma-finiteness are not required. Its measurability requirement on the limit is explicit in the form used here. The theorem is a standard result; its proof by truncation and domination is developed later in the course.
Because there are countably many functions, the separate exceptional null sets in the domination assumptions can be combined into one measurable null set: take their countable union. Thus the bounds hold simultaneously off a single null set. This is often useful when testing an alleged common majorant. Measurability of \(f\), by contrast, must not be omitted on an arbitrary measure space.
Two Preliminary Tests
The first test is to establish measurability of a pointwise limit when convergence holds everywhere. This does not settle the almost-everywhere case on an incomplete space, but it provides a useful direct check.
Proof. Fix \(a\in\mathbb{R}\). Since the rationals are dense in \(\mathbb{R}\), the following set identity holds: \[ \{x:f(x)>a\} = \bigcup_{\substack{q\in\mathbb{Q}\\q>a}} \ \bigcup_{N=1}^{\infty}\ \bigcap_{n\ge N}\{x:f_n(x)>q\}. \] Indeed, if \(f(x)>a\), choose a rational \(q\) with \(a<q<f(x)\). Convergence to \(f(x)\) implies that \(f_n(x)>q\) for all sufficiently large \(n\), so \(x\) belongs to the right side. Conversely, if \(f_n(x)>q\) for every \(n\ge N\), then convergence implies \(f(x)\ge q>a\), so \(x\) belongs to the left side. Each set \(\{f_n>q\}\) is measurable, and the right side uses only countable unions and intersections. Therefore \(\{f>a\}\in\mathcal{M}\) for every real \(a\), which proves that \(f\) is measurable. \(\square\)
The second test isolates exactly what “one integrable bound for the whole sequence” means. Here \(\sup_n|f_n|\) is the pointwise supremum, allowed to take the value \(+\infty\).
Proof. The function \(H\) is measurable because \(\{H>a\}=\bigcup_n\{|f_n|>a\}\) for every \(a\ge0\). If \(H\) is integrable, choose \(g=H\); then \(|f_n|\le g\) everywhere for every \(n\). Conversely, suppose such an integrable \(g\) exists. For each \(n\), choose a measurable null set \(N_n\) outside which \(|f_n|\le g\). The union \(N=\bigcup_nN_n\) is measurable and null by countable subadditivity. For \(x\notin N\), taking the supremum over \(n\) gives \(H(x)\le g(x)\). Monotonicity of the nonnegative Lebesgue integral now gives \(\int H\,d\mu\le\int g\,d\mu<\infty\), so \(H\) is integrable. \(\square\)
This test can disprove the existence of a common integrable majorant without guessing a candidate. It does not, by itself, imply that the integrals converge: the pointwise convergence and measurability hypotheses of the Dominated Convergence Theorem must still be checked.
Worked Examples: Apply the Tests
Worked Example: A Valid Interchange on a Finite Interval
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n(x)=x^n\) for \(n\ge1\). Determine the pointwise limit and compare its integral with the limit of the integrals.
Each \(f_n\) is measurable. If \(0\le x<1\), then \(x^n\to0\), while \(f_n(1)=1\) for every \(n\). Thus the pointwise limit is \(f=\chi_{\{1\}}\), which is measurable and has integral \(0\), since \(m(\{1\})=0\). For every \(x\in[0,1]\), \(0\le x^n\le1\), and the constant function \(g(x)=1\) is integrable on this finite-measure space, with \(\int_0^1 1\,dm=1\). The hypotheses of the Dominated Convergence Theorem hold.
Check: for \(n=1,2,3\), the integrals are \(1/2,1/3,1/4\), respectively, and these values tend to \(0\). The endpoint \(x=1\) does not change the integral because its singleton has Lebesgue measure zero.
Worked Example: Bounded Functions with No Integrable Bound on the Space
On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(f_n=\chi_{[n,n+1)}\). Test whether the Dominated Convergence Theorem justifies passing the limit through the integral.
Each \(f_n\) is measurable. For each fixed \(x\in\mathbb{R}\), there is at most one positive integer \(n\) for which \(x\in[n,n+1)\). Hence \(f_n(x)=0\) for all sufficiently large \(n\), and \(f_n(x)\to0\) everywhere. Direct calculation gives \[ \int_{\mathbb{R}}f_n\,dm=m([n,n+1))=1 \] for every \(n\), whereas \(\int_{\mathbb{R}}0\,dm=0\). Thus the interchange is false for this sequence.
Each function is bounded by \(1\), but \(1\) is not an integrable function on \(\mathbb{R}\): its integral is \(+\infty\). In fact, no common integrable majorant exists. If \(g\) were one, then for every \(n\), \(g\ge1\) \(m\)-a.e. on \([n,n+1)\). These intervals are pairwise disjoint, so countable additivity and monotonicity give \[ \int_{\mathbb{R}}g\,dm \ \ge\ \sum_{n=1}^{\infty}\int_{[n,n+1)}g\,dm \ \ge\ \sum_{n=1}^{\infty}1 \ =+\infty, \] contrary to integrability. Check: the pointwise limit has integral \(0\), while every integral in the sequence equals \(1\).
Worked Example: An Almost-Everywhere Limit That Is Not Measurable
This example shows why the limit’s measurability must be checked on a possibly incomplete space. Let \(X=[0,1]\), let \(\mathcal{M}=\mathcal{B}_{[0,1]}\), the Borel sigma-algebra, and let \(\mu\) be Lebesgue measure restricted to the Borel sets. Let \(C\) be the Cantor set, and choose a subset \(A\subset C\) that is not Borel. Such subsets exist because the Cantor set has non-Borel subsets. Define \(f_n=0\) everywhere and \(f=\chi_A\).
Every \(f_n\) is measurable. The sequence converges to \(f\) at every point outside \(C\), and \(\mu(C)=0\), so \(f_n\to f\) \(\mu\)-a.e. Also \(|f_n|\le g\) everywhere with \(g=0\), and \(\int g\,d\mu=0\). Nevertheless \(f\) is not measurable: the set \(\{x:f(x)>1/2\}\) is \(A\), which is not in \(\mathcal{B}_{[0,1]}\). Consequently \(\int f\,d\mu\) is not defined as a Lebesgue integral on this measurable space.
This does not contradict the Dominated Convergence Theorem: its stated hypotheses include measurability of \(f\). The example also illustrates the role of completeness. The measure space here is not complete, since it contains the Borel null set \(C\) but not all of its subsets. Check: the sequence integrals are all \(0\), but one cannot equate their limit to an integral of \(f\) in this space, because \(f\) is not measurable.
What Fails When a Hypothesis Is Missing?
No pointwise convergence, no identified limit. On \([0,1]\) with Lebesgue measure, set \(f_n(x)=(-1)^n\). These functions are measurable and \(\int_0^1f_n\,dm=(-1)^n\), which does not converge. At every point, \(f_n(x)\) also fails to converge. There is therefore no pointwise limit to integrate, and the hypotheses of the Dominated Convergence Theorem are not met. A uniform bound alone cannot repair the absence of convergence.
A common bound need not be integrable. A frequent incorrect argument for the escaping-interval example is: “\(|f_n|\le1\), so apply dominated convergence.” The inequality is true, but on \((\mathbb{R},\mathcal{L},m)\) the proposed majorant has integral \[ \int_{\mathbb{R}}1\,dm=+\infty, \] not a finite value. The correction is to test the integral of the majorant on the actual measure space, not merely its pointwise size. In the example, the integrals are \(1\) for all \(n\) while the integral of the limit is \(0\).
Convergence theorems give different conclusions. The Monotone Convergence Theorem applies to measurable functions \(f_n:X\to[0,\infty]\) on a measure space when \(f_n\uparrow f\) pointwise; it gives \(\int f_n\,d\mu\to\int f\,d\mu\), with \(+\infty\) allowed. It does not apply merely because the functions are nonnegative. Fatou's lemma, for nonnegative measurable \(f_n\) on a measure space, gives only \[ \int\liminf_{n\to\infty}f_n\,d\mu \le \liminf_{n\to\infty}\int f_n\,d\mu. \] That inequality is not an equality claim. For the escaping-interval example, its left side is \(0\) and its right side is \(1\), so the inequality holds but does not justify interchanging limit and integral.
When auditing a proof, write down the exact theorem and match each assumption to a fact about the sequence. In particular, check whether the limit is measurable, whether convergence is pointwise \(\mu\)-a.e. rather than merely suggestive from a graph or a numerical pattern, and whether the proposed majorant has finite integral. If any check fails, the proposed interchange has not been justified; a counterexample may show that the identity itself is false, or the integral may not even be defined.
Check Your Understanding
For each question, identify which hypotheses are relevant before deciding what conclusion is valid.
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\). What is the pointwise limit, and what integrable function dominates every \(f_n\)?
- On \(\mathbb{R}\) with Lebesgue measure, let \(f_n=\chi_{[n,n+1)}\). Which Dominated Convergence hypothesis fails, and what are the two sides of the proposed interchange?
- Why does measurability of every \(f_n\) not, on an incomplete measure space, guarantee measurability of a proposed limit that agrees with the pointwise limit only almost everywhere?
- For a measurable sequence \(f_n\), explain why integrability of \(H=\sup_n|f_n|\) is equivalent to existence of a common nonnegative integrable majorant.
- What conclusion does Fatou's lemma give for nonnegative measurable functions, and why does that conclusion not in general assert equality?