Why Bounds Need Only Be Local
For an integral depending on a parameter, continuity is a local question: to understand the integral near one parameter value, only parameter values near that point matter. A dominating function that works throughout the entire parameter domain may therefore be unnecessary. It is enough that every compact region of parameters has its own common integrable bound. This permits the bound to become larger as the compact region moves through the domain.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Let \(U\) be an open subset of \(\mathbb{R}^d\), and let \(f:X\times U\to\mathbb{C}\) be such that \(x\mapsto f(x,t)\) is measurable for each \(t\in U\). Write \(f_t(x)=f(x,t)\). A complex-valued measurable function is integrable when \(\int_X |f_t|\,d\mu<\infty\). The integral of a complex-valued function is defined by integrating its real and imaginary parts.
Since a compact set is not required to be a neighborhood of any particular point, the condition is especially useful when stated for every compact \(K\subset U\). In applying a convergence theorem to a sequence of parameters, the possibly different exceptional sets cause no difficulty: a sequence uses only countably many parameters, and a countable union of null sets is null.
The Local Domination Criterion
Proof. Fix \(t\in U\). Since \(U\) is open, there is a closed ball \(K\subset U\) containing \(t\). The assumed bound gives \(|f_t|\le g_K\) \(\mu\)-a.e., so \(\int_X|f_t|\,d\mu\le\int_Xg_K\,d\mu<\infty\). Thus \(F(t)\) exists and is finite.
Now fix \(t\in U\) and consider any sequence \(t_n\to t\) in \(U\). Choose a closed ball \(K\subset U\) centered at \(t\) with positive radius. There is an \(N\) such that \(t_n\in K\) whenever \(n\ge N\). For each such \(n\), \(|f_{t_n}|\le g_K\) outside a null set \(N_n\); also \(|f_t|\le g_K\) outside a null set. The union of these countably many exceptional sets is null. Outside that union, the parameter continuity assumption gives \(f(x,t_n)\to f(x,t)\), and the bound by \(g_K\) holds for every \(n\ge N\). The Dominated Convergence Theorem applies to the tail sequence: its functions are measurable, converge \(\mu\)-a.e., and are dominated in absolute value by the integrable function \(g_K\). Therefore \(\int_X f_{t_n}\,d\mu\to\int_X f_t\,d\mu\) as \(n\to\infty\). Omitting finitely many terms does not change this limit. Since this holds for every sequence \(t_n\to t\), the sequential criterion for continuity in the metric space \(U\subset\mathbb{R}^d\) proves that \(F\) is continuous at \(t\). The choice of \(t\) was arbitrary. \(\square\)
The theorem does not require the bound \(g_K\) to be the same for different compact sets. Nor does it require convergence of the integrands to be uniform in \(x\). It uses continuity in the parameter for almost every \(x\), together with domination for the sequence of parameter values being considered.
Proof. Let \(K\subset U\) be compact. The neighborhoods \(V_{t_0}\), as \(t_0\) ranges over \(K\), form an open cover of \(K\). Compactness gives a finite subcover \(V_{t_1},\ldots,V_{t_r}\). Set \(g_K=\sum_{j=1}^r g_{t_j}\). This function is nonnegative and integrable because it is a finite sum of integrable functions. For every \(s\in K\), some \(V_{t_j}\) contains \(s\), so \(|f_s|\le g_{t_j}\le g_K\) \(\mu\)-a.e. Thus the compact-set hypothesis of the theorem holds, and the theorem proves continuity. \(\square\)
Worked Examples of Compact-Local Bounds
Worked Example: An Exponential Family on an Infinite Measure Space
Take \(X=(0,\infty)\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\), let \(U=(0,\infty)\), and set \(f(x,t)=e^{-tx}\). For a compact parameter interval \(K=[a,b]\subset(0,\infty)\), we have \(a>0\). Since \(x>0\) and \(t\ge a\), \(e^{-tx}\le e^{-ax}\). Thus \(g_K(x)=e^{-ax}\) is a common bound, and
For each \(x>0\), the function \(t\mapsto e^{-tx}\) is continuous. The theorem applies on each compact interval, so \(F(t)=\int_0^\infty e^{-tx}\,dx\) is continuous for \(t>0\). Direct integration gives \(F(t)=1/t\): the antiderivative is \(-e^{-tx}/t\), whose value tends to \(0\) at infinity and equals \(-1/t\) at \(x=0\). Check: at \(t=2\), the formula gives \(F(2)=1/2\); integrating \(e^{-2x}\) gives \([-\tfrac12e^{-2x}]_0^\infty=1/2\).
Worked Example: Powers Near the Integrability Threshold
Take \(X=(0,1)\) with Lebesgue measure \(m\), \(U=(-1,\infty)\), and \(f(x,t)=x^t\). For a compact interval \(K=[a,b]\subset U\), its left endpoint satisfies \(a>-1\). If \(a<0\), then \(t\ge a\) and \(0<x<1\) imply \(x^t\le x^a\); choose \(g_K(x)=x^a\). Its integral is \(\int_0^1x^a\,dx=1/(a+1)<\infty\). If \(a\ge0\), then \(x^t\le1\); choose \(g_K(x)=1\), whose integral over \((0,1)\) is \(1\). The parameter dependence is continuous for each \(x\in(0,1)\), so the theorem gives continuity of \(F\) on \(U\).
For every \(t>-1\), direct integration yields \(\displaystyle F(t)=\int_0^1x^t\,dx=[x^{t+1}/(t+1)]_0^1=1/(t+1)\). For example, on \(K=[-1/2,1]\) a common bound is \(x^{-1/2}\), and its integral is \(1/(-1/2+1)=2\). At \(t=1/2\), the integral is \([x^{3/2}/(3/2)]_0^1=2/3\), agreeing with \(1/(1+1/2)=2/3\). The bound is integrable because the compact interval stays above the threshold \(t=-1\).
Worked Example: A Geometric Series Under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure \(\mu\), take \(U=(-1,1)\), and define \(f(k,t)=t^k\). For a compact \(K\subset(-1,1)\), let \(q=\max_{t\in K}|t|\). Compactness and \(K\subset(-1,1)\) imply \(0\le q<1\). For all \(t\in K\), \(|f(k,t)|=|t|^k\le q^k\). The function \(g_K(k)=q^k\) is integrable for counting measure, since
When \(q=0\), every term is zero and the same formula gives zero. For each fixed \(k\), \(t\mapsto t^k\) is continuous. Therefore the theorem proves continuity of \(F(t)=\sum_{k=1}^{\infty}t^k\) on \((-1,1)\). Summing the geometric series gives \(F(t)=t/(1-t)\). Check: at \(t=-1/2\), the series is \(-1/2+1/4-1/8+\cdots\), whose geometric sum is \((-1/2)/(1-(-1/2))=-1/3\); the formula gives the same value.
A Common Pitfall: Demanding One Bound on the Whole Domain
For the exponential example, it would be incorrect to claim that \(e^{-x}\) dominates \(e^{-tx}\) for every \(t>0\) and \(x>0\). At \(t=1/2\) and \(x=1\), the proposed inequality would say \(e^{-1/2}\le e^{-1}\), which is false. The correction is to choose a bound for the compact parameter interval under consideration: on \(K=[a,b]\subset(0,\infty)\), use \(e^{-ax}\), as verified above.
In fact, the failure of this particular global bound is not a failure of the continuity argument. The integral remains \(F(t)=1/t\), continuous throughout \((0,\infty)\). Local domination asks for a suitable integrable function on each compact part of the parameter domain, not for a single integrable function that works everywhere. A bound may legitimately depend on the compact set; the theorem’s conclusion follows by applying dominated convergence near each parameter value.
Check Your Understanding
For each question, identify the parameter domain and verify that the proposed bound is integrable before applying the continuity criterion.
- For \(f(x,t)=e^{-tx}\) on \((0,\infty)\), what bound works on a compact interval \([a,b]\subset(0,\infty)\), and why is it integrable?
- For \(f(x,t)=x^t\) on \((0,1)\), explain why the condition \(t>-1\) is relevant to the local bound near \(x=0\).
- For \(f(k,t)=t^k\) under counting measure, find a summable bound on a compact \(K\subset(-1,1)\) and state why its ratio is less than one.
- In the theorem’s proof, why is it enough to have a bound outside an exceptional null set that may depend on the parameter?
- Does continuity of \(F(t)=\int_X f(x,t)\,d\mu(x)\) require one integrable bound valid on all of \(U\)? Explain the role of compact-local bounds.