Why a Common Integrable Bound Matters
Pointwise convergence controls the value of a function at each fixed point, but it does not by itself control how much mass the functions carry across the whole space. The Dominated Convergence Theorem gives a precise condition under which pointwise almost-everywhere convergence does control the integrals: every function in the sequence must be bounded in absolute value by the same integrable function. In this tutorial we state the theorem and examine what its hypotheses do. Its proof is given in the following tutorial.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space; it need not be finite, sigma-finite, or complete. Functions \(f_n,f,g:X\to\mathbb{R}\) or \(\mathbb{C}\) are measurable and finite-valued, except that a majorant may be extended-real-valued. Integrability means \(\int |f|\,d\mu<\infty\), with the absolute value interpreted as complex modulus when appropriate. Almost-everywhere statements mean that the set where the assertion fails is contained in a measurable set of measure zero. We assume the limit function \(f\) is measurable, which matters on a measure space that is not complete.
The bound is required only almost everywhere for each \(n\); it need not hold on the same exceptional set initially. Since there are countably many functions, the union of their exceptional null sets is null. Thus there is one null set outside which all the bounds hold. Likewise, convergence need only hold almost everywhere, not at every point. The majorant \(g\), however, must be integrable: a finite bound at each point is not enough, especially on an infinite-measure space.
What the Hypotheses Give
Proof. For each \(n\), choose a measurable null set \(N_n\) outside which \(|f_n|\le g\), and choose a measurable null set \(N_0\) outside which \(f_n(x)\to f(x)\). The union \(N=N_0\cup\bigcup_{n=1}^{\infty}N_n\) is measurable and null by countable subadditivity. If \(x\notin N\), then \(|f_n(x)|\le g(x)\) for every \(n\), and \(f_n(x)\to f(x)\). If \(g(x)<\infty\), continuity of absolute value gives \(|f(x)|=\lim_n|f_n(x)|\le g(x)\). If \(g(x)=\infty\), the inequality \(|f(x)|\le g(x)\) also holds. Hence \(|f|\le g\) outside \(N\). Monotonicity of the nonnegative Lebesgue integral, and the fact that changing an integrand on a null set does not change its integral, give \(\int|f|\,d\mu\le\int g\,d\mu<\infty\). Therefore \(f\) is integrable. \(\square\)
This proposition explains why the same majorant controls both the sequence and its limit. The theorem’s conclusion is stronger than convergence of the signed or complex integrals: the integral of the absolute difference tends to zero. That conclusion is often called convergence in \(L^1(\mu)\).
Proof. The integral triangle inequality applies because \(f_n\) and \(f\) are integrable. It gives \(\left|\int(f_n-f)\,d\mu\right|\le\int|f_n-f|\,d\mu\), and linearity identifies the left side with \(\left|\int f_n\,d\mu-\int f\,d\mu\right|\). The Dominated Convergence Theorem says that the right side tends to zero. Thus the difference of the integrals tends to zero, as claimed. \(\square\)
Worked Examples
Worked Example: Powers on a Finite Interval
On \(([0,1],\mathcal{L},m)\), let \(f_n(x)=x^n\) and determine the limit of their integrals. The functions are measurable, and for \(0\le x<1\), \(x^n\to0\); at \(x=1\), \(x^n=1\) for every \(n\). Thus the pointwise limit is \(f=\chi_{\{1\}}\), which is zero \(m\)-a.e. The common bound is \(g(x)=1\): for every \(x\in[0,1]\) and every \(n\ge1\), \(0\le x^n\le1\). This bound is integrable, since \(\int_0^1 1\,dm=1\).
Direct computation gives $$ \int_0^1 x^n\,dm=\left[\frac{x^{n+1}}{n+1}\right]_{0}^{1}=\frac{1}{n+1}. $$ Therefore \(\lim_{n\to\infty}\int_0^1x^n\,dm=0\). The limit function has integral \(\int_0^1\chi_{\{1\}}\,dm=m(\{1\})=0\), agreeing with the theorem. Check: at \(n=1\), the integral is \(1/(1+1)=1/2\); at \(n=2\), it is \(1/(2+1)=1/3\); both are consistent with \(1/(n+1)\to0\). The single point where the pointwise limit differs from zero has measure zero, so it does not change the integral.
Worked Example: A Sequence on Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with its full sigma-algebra and counting measure \(\mu\), so that \(\int h\,d\mu=\sum_{k=1}^{\infty}h(k)\) for nonnegative \(h\). Define \(f_n(k)=1\) if \(k=n\), and \(f_n(k)=0\) otherwise. For each fixed \(k\), \(f_n(k)=0\) for every \(n>k\), so \(f_n(k)\to0\) everywhere. Also, \(|f_n(k)|\le g(k)=1\) for all \(n,k\).
But this bound is not integrable: \(\int g\,d\mu=\sum_{k=1}^{\infty}1=\infty\). Each function has integral \(\int f_n\,d\mu=1\), because it equals one at exactly one point and zero elsewhere, while the limit function has integral zero. Consequently, the integrals do not converge to the integral of the limit. Check: the partial sum of \(g\) over \(k=1,\ldots,K\) is \(K\), which is unbounded as \(K\to\infty\); thus \(g\) is indeed not integrable. This example has everywhere convergence and a common finite pointwise bound, but it lacks an integrable majorant.
Worked Example: A Bounded Sequence Without a Limit
Take the one-point measure space \(X=\{a\}\), with \(\mathcal{M}=\{\varnothing,\{a\}\}\) and \(\mu(\{a\})=1\). Define \(f_n(a)=(-1)^n\). These functions are measurable and satisfy \(|f_n(a)|=1\), so \(g(a)=1\) is an integrable common bound: \(\int g\,d\mu=1\). However, \(f_n(a)\) does not converge. The even-indexed terms equal \(1\), while the odd-indexed terms equal \(-1\), and these two subsequences have different limits.
The integrals are \(\int f_n\,d\mu=f_n(a)=(-1)^n\), so they alternate between \(-1\) and \(1\) and have no limit. Check: the measure of the only point is one, so integrating a constant \(c\) gives \(c\cdot1=c\); hence the displayed integral values follow directly. Here the common integrable bound holds, but almost-everywhere convergence fails (on this space, the point \(a\) is not a null set). A bound alone does not permit passage of a limit through the integral.
Common Mistakes and the Role of Each Condition
Mistake: pointwise boundedness is treated as domination. On \(([0,1],\mathcal{L},m)\), set \(f_n=n\chi_{(0,1/n)}\). For every fixed \(x\in[0,1]\), \(f_n(x)\to0\): at \(x=0\) every value is zero, and for \(x>0\), \(f_n(x)=0\) once \(n\ge1/x\). Nevertheless, $$ \int_0^1 f_n\,dm=n\cdot m((0,1/n))=n\cdot\frac1n=1. $$ Thus the integrals do not converge to the integral of the zero limit.
The error would be to assume a common integrable majorant just because each \(f_n\) is individually bounded. To see why no such majorant exists, suppose \(g\) were integrable and \(g\ge f_n\) a.e. for every \(n\). Removing the countable union of exceptional null sets, the inequalities would hold simultaneously. For each \(x\in(0,1/4)\), choose \(n=\lfloor1/(2x)\rfloor\). Then \(n\ge1/(4x)\) and \(nx\le1/2<1\), so \(x\in(0,1/n)\) and \(f_n(x)=n\ge1/(4x)\). Hence \(g(x)\ge1/(4x)\) for almost every \(x\in(0,1/4)\), which contradicts integrability because \(\int_0^{1/4}(1/(4x))\,dm=\infty\). The integrals \(1\) in this calculation are exact: the support has length \(1/n\) and the function’s value there is \(n\).
The conditions should be checked separately rather than blended together:
- Measurability: the functions and the limit must be measurable for their Lebesgue integrals to be covered by the theorem.
- Almost-everywhere convergence: convergence may fail on a null set, but it cannot simply be omitted; the one-point example shows the failure when the sequence does not converge.
- A single common bound: the function \(g\) must bound every \(f_n\), not merely each one with a different bound.
- Integrability of the bound: \(g\) must satisfy \(\int g\,d\mu<\infty\), not just be finite at each point. The counting-measure example shows why.
- The conclusion concerns integrals: the theorem gives convergence of the integrals and, more strongly, convergence of \(\int|f_n-f|\,d\mu\) to zero. It does not assert uniform convergence of \(f_n\) to \(f\).
These distinctions are especially useful on spaces of infinite measure, where a constant pointwise bound may fail to be integrable. On finite-measure spaces, a constant bound is integrable, but the sequence must still converge almost everywhere. In every application, identify the measure space, verify measurability, check the mode of convergence, and exhibit one integrable function that dominates the entire sequence almost everywhere.
Check Your Understanding
For each question, identify which hypotheses apply and justify any use of the theorem.
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\). State the almost-everywhere limit and an integrable common bound, then compute the limiting integral.
- On \(\mathbb{N}\) with counting measure, why does \(g(k)=1\) fail to be an integrable majorant, even though it is finite at every \(k\)?
- On a one-point space of measure one, let \(f_n(a)=(-1)^n\). Which hypothesis fails, and what are the integrals?
- For \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\), compute each integral and explain why the sequence cannot have an integrable common majorant.
- Under the Dominated Convergence Theorem’s hypotheses, what stronger conclusion about \(\int|f_n-f|\,d\mu\) accompanies convergence of the integrals?