Continuity from a Local Integrable Majorant
An integral can depend continuously on a parameter even when the integrand varies substantially across the underlying space. The key is to control all nearby integrands by one integrable function, rather than to control each parameter value separately. This gives a direct application of the Dominated Convergence Theorem to every sequence of parameter values approaching the point of interest.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite or sigma-finite. Let \(I\subseteq\mathbb{R}\) be an interval, and let \(f:I\times X\to\mathbb{R}\) be real-valued. Assume that for each \(t\in I\), the function \(x\mapsto f(t,x)\) is \(\mathcal{M}\)-measurable. Its integral is a finite real number whenever it is dominated almost everywhere by an integrable function. The continuity considered below is continuity of the real-valued function \(F(t)=\int_X f(t,x)\,d\mu(x)\).
- for every \(x\in X\setminus N\), the function \(t\mapsto f(t,x)\) is continuous at \(t_0\), relative to \(I\); and
- for every \(x\in X\setminus N\) and every \(t\in I\) with \(|t-t_0|<\delta\), \(|f(t,x)|\le g(x)\).
The same null set is used for continuity and domination; changing values on that null set does not affect the integrals. The bound is local in the parameter: it need only hold in a neighborhood of the point under consideration. We use the Dominated Convergence Theorem in its standard sequential form: if measurable real-valued functions \(u_n\) converge \(\mu\)-a.e. to \(u\) and \(|u_n|\le g\) \(\mu\)-a.e. for one integrable \(g\), then \(u\) is integrable and \(\int u_n\,d\mu\to\int u\,d\mu\). That theorem, including its proof, belongs to the earlier material on dominated convergence.
Proof. First, for every \(t\) in the stated neighborhood, \(|f(t,x)|\le g(x)\) for \(\mu\)-a.e. \(x\). Thus \(f(t,\cdot)\) is integrable and \(F(t)\) is finite. To prove continuity at \(t_0\), let \((t_n)\) be any sequence in \(I\) with \(t_n\to t_0\). There is an index \(n_0\) such that \(|t_n-t_0|<\delta\) for all \(n\ge n_0\). For each \(x\notin N\), continuity in the parameter gives \(f(t_n,x)\to f(t_0,x)\). For all \(n\ge n_0\), the same integrable \(g\) dominates \(|f(t_n,x)|\) outside \(N\). The Dominated Convergence Theorem therefore gives \(F(t_n)\to F(t_0)\) for the tail \(n\ge n_0\), and hence for the full sequence. In the real line, sequential continuity is equivalent to continuity: if \(F\) were not continuous at \(t_0\), there would be an \(\varepsilon>0\) such that for every \(n\) one could choose \(t_n\in I\) with \(|t_n-t_0|<1/n\) but \(|F(t_n)-F(t_0)|\ge\varepsilon\), contradicting the convergence just proved. Thus \(F\) is continuous at \(t_0\). \(\square\)
This argument proves a result at one point at a time. Its quantifiers matter: the majorant must be a single integrable \(g\) for all nearby parameter values, up to a null set. It is not enough that each individual integrand is integrable, or even that each has some integrable majorant that depends on the parameter.
Worked Example: An Exponential Integral on the Half-Line
Let \(X=[0,\infty)\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\). For \(t>0\), set \(f(t,x)=e^{-tx}\). Show that \(F(t)=\int_0^\infty e^{-tx}\,dx\) is continuous at \(t_0=2\).
For each fixed \(x\ge0\), \(t\mapsto e^{-tx}\) is continuous. If \(|t-2|<1\), then \(1<t<3\), so \(tx\ge x\) and \(0\le e^{-tx}\le e^{-x}\). The majorant is integrable, since \(\int_0^\infty e^{-x}\,dx=1\). The theorem applies. Direct integration also gives, for every \(t>0\),
For example, at \(t=2+\tfrac14=\tfrac94\), \(F(t)=\tfrac49\), while \(F(2)=\tfrac12\); their difference in absolute value is \(\left|\tfrac49-\tfrac12\right|=\tfrac1{18}\). The formula \(F(t)=1/t\) is continuous at \(2\), confirming the theorem's conclusion. Check: substituting \(t=2\) into the integral formula gives \(\int_0^\infty e^{-2x}\,dx=[-\tfrac12e^{-2x}]_0^\infty=\tfrac12\), as required.
Worked Example: A Power Integral Near Parameter Zero
Take \(X=(0,1)\) with Lebesgue measure and \(I=(-\tfrac12,\tfrac12)\). For \(t\in I\), define \(f(t,x)=x^t\). Show that the integral is continuous at \(t_0=0\).
For each \(x\in(0,1)\), \(t\mapsto x^t\) is continuous. Since \(t>-\tfrac12\) and \(0<x<1\), raising \(x\) to the larger exponent gives the smaller value, so \(0<x^t\le x^{-1/2}\). The function \(g(x)=x^{-1/2}\) is integrable because \(\int_0^1x^{-1/2}\,dx=2\). Thus the theorem applies, and direct evaluation yields
where \(t+1>\tfrac12\), so the lower endpoint term is zero. At \(t=\tfrac14\), the integral is \(1/(1+\tfrac14)=\tfrac45\); at \(t=0\), it is \(1\). The difference is \(\left|\tfrac45-1\right|=\tfrac15\). Check: the majorant has integral \(2\), and the formula at \(t=0\) gives \(F(0)=1=\int_0^1 1\,dx\).
Worked Example: A Continuously Varying Series
Let \(X=\mathbb{N}_0=\{0,1,2,\ldots\}\) with the counting sigma-algebra and counting measure \(\mu\). For \(-1<t<1\), define \(f(t,n)=(t/3)^n\), with the \(n=0\) term defined to be \(1\), including at \(t=0\). Prove that the sum \(F(t)=\int_X f(t,n)\,d\mu(n)\) is continuous at \(t_0=0\).
Every function on this countable space is measurable, and for each fixed \(n\), \(f(t,n)\) is continuous in \(t\). For \(|t|<1\), \(|f(t,n)|=|t/3|^n\le 3^{-n}\), including the \(n=0\) term. The majorant is integrable for counting measure because
Consequently the theorem applies. The geometric-series formula gives \(F(t)=\sum_{n=0}^\infty(t/3)^n=1/(1-t/3)\), since \(|t/3|<1\). At \(t=\tfrac12\), \(F(t)=1/(1-\tfrac16)=\tfrac65\), and at \(t=0\), \(F(0)=1\). Check: the first four terms at \(t=\tfrac12\) are \(1,\tfrac16,\tfrac1{36},\tfrac1{216}\); their sum is \(\tfrac{259}{216}\), while the exact sum is \(\tfrac65=\tfrac{259}{216}+\tfrac{1}{1080}\), with the positive remainder accounting for the omitted terms.
A Pointwise Limit Is Not Enough
The local majorant is what licenses the passage from convergence of integrands to convergence of their integrals. The following example shows that pointwise continuity alone does not suffice. It also makes the missing hypothesis concrete rather than merely formal.
Worked Example: A Narrow Spike with a Constant Integral
On \(X=(0,1)\) with Lebesgue measure, define \(f(0,x)=0\), and, for \(0<t<\tfrac12\), define \(f(t,x)=t^{-1}\chi_{(0,t)}(x)\). For each fixed \(x\in(0,1)\), \(f(t,x)\to0\) as \(t\downarrow0\): if \(x>0\), then \(f(t,x)=0\) whenever \(0<t\le x\). Thus the integrand is pointwise continuous at \(t=0\). But for every \(0<t<\tfrac12\),
So \(F\) is not continuous at zero. No integrable function can dominate all these nearby integrands, even if domination is only required almost everywhere for each parameter separately. Indeed, suppose an integrable \(g\ge0\) did so. For each rational \(r\in(0,\tfrac12)\), domination would imply \(g(x)\ge1/r\) for almost every \(x\in(0,r)\). Taking the union of the exceptional null sets over these countably many rationals still gives a null set. Outside it, for every \(x\in(0,\tfrac12)\) and every rational \(r\) with \(x<r<\tfrac12\), one has \(g(x)\ge1/r\). Letting such rationals decrease to \(x\) gives \(g(x)\ge1/x\). But \(\int_0^{1/2}x^{-1}\,dx=\infty\), contradicting integrability of \(g\). Check: the integral of each spike is exactly its height \(1/t\) times its interval length \(t\), hence is \(1\), not a quantity tending to zero.
Using the Criterion Correctly
In an application at \(t_0\), check the following in order:
- For every fixed \(t\) in a neighborhood, \(x\mapsto f(t,x)\) is measurable.
- For almost every \(x\), the parameter function \(t\mapsto f(t,x)\) is continuous at the specific point \(t_0\).
- One nonnegative measurable \(g\) has finite integral and bounds \(|f(t,x)|\) for every nearby \(t\), outside a null set.
A common incorrect calculation is to observe \(f(t,x)\to f(t_0,x)\) pointwise and immediately write \(\lim_{t\to t_0}\int f(t,x)\,d\mu(x)=\int f(t_0,x)\,d\mu(x)\). The spike example disproves that step: its pointwise limit is zero, but its integrals remain one. The correction is not to interchange the limit and integral without justification. Instead, apply the Dominated Convergence Theorem to an arbitrary parameter sequence and verify one integrable majorant works throughout a neighborhood. If that verification fails, the continuity conclusion has not been established.
In particular, the theorem does not require a finite measure space: its examples include the half-line with Lebesgue measure and the infinite counting space \(\mathbb{N}_0\). What matters is integrability of the majorant, not finiteness of the total measure. The conclusion concerns continuity of the integral as a function of the parameter; it asserts neither uniform convergence of the integrands nor differentiability of the integral.
Check Your Understanding
For each question, identify the pointwise convergence and the precise integrable majorant needed, or explain why the criterion cannot be applied.
- On \((0,1)\), consider \(f(t,x)=x^t\) for \(t\in(-\tfrac12,\tfrac12)\). What majorant proves continuity at \(t_0=\tfrac14\), and what is the integral at that parameter?
- On \([0,\infty)\), consider \(f(t,x)=e^{-tx}\) near \(t_0=3\). Give a neighborhood and an integrable majorant on it.
- For the counting-measure family \(f(t,n)=(t/3)^n\), compute the integral at \(t=-\tfrac12\) and state why the same majorant applies.
- For the narrow-spike family, compute \(F(t)\) for \(0<t<\tfrac12\). Which hypothesis of the continuity theorem fails?
- Explain why having a different integrable majorant for each individual parameter does not, by itself, verify the theorem's local domination hypothesis.