From Difference Quotients to Derivatives of Integrals
Suppose a parameter-dependent function \(f(t,x)\) is integrated over \(x\), giving \(F(t)=\int_X f(t,x)\,d\mu(x)\). To find \(F'(t_0)\), the natural candidate is \(\int_X \partial_t f(t_0,x)\,d\mu(x)\). The relevant question is whether the difference quotients of \(f\) converge in a way that permits passage of the limit through the integral. Dominated Convergence gives a criterion: pointwise convergence of the quotients is enough when they share one integrable bound.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, not assumed finite or sigma-finite. The functions in the criterion are real-valued and measurable in \(x\) for each parameter \(t\). Integrability means Lebesgue integrability, or equivalently \(\int_X |f(t,x)|\,d\mu(x)<\infty\). The derivative in the parameter is an ordinary two-sided derivative.
Proof. First fix \(t\) with \(|t-t_0|<\delta\). If \(t=t_0\), integrability is assumed. Otherwise set \(h=t-t_0\). The bound gives $$ |f(t,x)|\le |f(t_0,x)|+|h|g(x) $$ for every \(x\). The right side is integrable, so \(f(t,\cdot)\) is integrable by comparison.
Write \(q_h(x)=(f(t_0+h,x)-f(t_0,x))/h\) for \(0<|h|<\delta\). Each \(q_h\) is measurable because both sections in its numerator are measurable. The derivative \(d(x)=\partial_t f(t_0,x)\) is measurable: it is the pointwise limit of the measurable functions \(q_{\delta/(n+1)}\), for \(n\ge1\). Since \(|q_{\delta/(n+1)}|\le g\), taking the pointwise limit gives \(|d|\le g\), so \(d\) is integrable.
Linearity of the integral for integrable functions gives $$ \frac{F(t_0+h)-F(t_0)}{h}=\int_X q_h\,d\mu. $$ To prove that the right side tends to \(\int_X d\,d\mu\), take any sequence of nonzero \(h_n\to0\) with \(|h_n|<\delta\). By the assumed pointwise derivative, \(q_{h_n}(x)\to d(x)\) for every \(x\). The functions \(q_{h_n}\) are measurable and satisfy \(|q_{h_n}|\le g\), where \(g\) is integrable. The Dominated Convergence Theorem therefore gives \(\int_X q_{h_n}\,d\mu\to\int_X d\,d\mu\). This holds for every such sequence. In particular, if the difference quotient of \(F\) did not tend to \(\int_X d\,d\mu\), there would be an \(\varepsilon>0\) and a sequence \(h_n\to0\) for which its distance from that value is at least \(\varepsilon\), contradicting the convergence just proved. Thus \(F'(t_0)=\int_X d\,d\mu\). \(\square\)
How the Mean Value Theorem Supplies Domination
Proof. Fix \(x\) and \(h\) with \(0<|h|<\delta\). The scalar mean value theorem on the interval with endpoints \(t_0\) and \(t_0+h\) gives a point \(s\) strictly between them such that $$ \left|\frac{f(t_0+h,x)-f(t_0,x)}{h}\right| =|\partial_t f(s,x)| \le G(x). $$ The point \(s\) lies in \((t_0-\delta,t_0+\delta)\), so the stated derivative bound applies. This holds for every \(x\) and every permitted \(h\), which is exactly the required common integrable bound. \(\square\)
The bound must be independent of \(h\). A bound that is integrable for each fixed quotient but grows without control as \(h\to0\) does not provide the hypothesis of Dominated Convergence.
Worked Example: Differentiating an Exponential Integral on a Bounded Interval
Let \(X=[0,1]\) with Lebesgue measure \(m\), and set \(f(t,x)=e^{tx}\) for real \(t\). Compute the derivative at \(t_0=0\) of \(F(t)=\int_0^1 e^{tx}\,dx\).
For \(|h|<1\), the mean value theorem applied to \(s\mapsto e^{sx}\) gives, for some \(s\) between \(0\) and \(h\), $$ \left|\frac{e^{hx}-1}{h}\right|=x e^{sx}\le e x, $$ because \(0\le x\le1\) and \(|s|<1\). The function \(g(x)=ex\) is integrable on \([0,1]\), with \(\int_0^1 ex\,dm=e/2\). Also \(f(0,x)=1\) is integrable, and each section is continuous, hence measurable. The pointwise derivative at zero is \(\partial_t f(0,x)=x\). The criterion therefore yields $$ F'(0)=\int_0^1x\,dm=\left[\frac{x^2}{2}\right]_0^1=\frac12. $$
Check by direct calculation. For \(t\ne0\), integration gives \(F(t)=(e^t-1)/t\), while \(F(0)=1\). Hence $$ \frac{F(h)-F(0)}{h}=\frac{e^h-1-h}{h^2}\longrightarrow\frac12 $$ by the second-order Taylor expansion of \(e^h\) at zero. This agrees with the value obtained from the difference-quotient criterion.
Worked Example: A Parameter Integral on an Infinite Measure Space
Let \(X=(0,\infty)\) with Lebesgue measure and \(f(t,x)=e^{-x}\sin(tx)\), for \(t\in\mathbb{R}\). Find \(F'(0)\), where \(F(t)=\int_0^\infty f(t,x)\,dx\).
For each fixed \(t\), \(|f(t,x)|\le e^{-x}\), so the section is integrable. Also \(f(0,x)=0\). For \(h\ne0\), the mean value theorem applied to \(s\mapsto\sin(sx)\) gives \(|\sin(hx)/h|\le x\), because the derivative is \(x\cos(sx)\) and \(|\cos(sx)|\le1\). Thus $$ \left|\frac{f(h,x)-f(0,x)}{h}\right|\le xe^{-x}. $$ The proposed dominator is integrable: integration by parts gives $$ \int_0^\infty xe^{-x}\,dx =\left[-(x+1)e^{-x}\right]_0^\infty=1. $$ The pointwise derivative at zero is \(xe^{-x}\). All sections are measurable, so the criterion applies and gives \(F'(0)=1\).
Independent check. For \(t\in\mathbb{R}\), put \(I(t)=\int_0^\infty e^{-x}\sin(tx)\,dx\) and \(J(t)=\int_0^\infty e^{-x}\cos(tx)\,dx\). These integrals converge absolutely. Integration by parts, with boundary terms vanishing at infinity because \(e^{-x}\) tends to zero, gives \(I(t)=tJ(t)\) and \(J(t)=1-tI(t)\). Consequently \(I(t)=t(1-tI(t))\), so \(F(t)=I(t)=t/(1+t^2)\). Therefore \(F'(0)=1\), as above.
Why Pointwise Differentiability Is Not Enough
A pointwise derivative describes what happens at each fixed \(x\), but does not prevent the difference quotients from becoming large on sets whose locations or sizes change with \(h\). Without an integrable bound, the integral of a quotient can fail to approach the integral of its pointwise limit.
Worked Example: A Two-Sided Pointwise Derivative but a Discontinuous Integral
Take \(X=(0,1)\) with Lebesgue measure. Define \(f(t,x)=0\) for \(t\le0\), and for \(t>0\) define $$ f(t,x)=\frac{1}{t}\chi_{(0,\min(t,1))}(x). $$ Every section is measurable and integrable. At \(t_0=0\), for each fixed \(x\in(0,1)\), \(f(t,x)=0\) whenever \(|t|<x\): for positive such \(t\), \(x\notin(0,t)\), and for nonpositive \(t\), the function is zero by definition. Thus the ordinary two-sided pointwise derivative at zero exists and equals zero for every \(x\).
However, for \(0<t<1\), $$ F(t)=\int_0^1 \frac1t\chi_{(0,t)}(x)\,dx =\frac1t\,m((0,t))=\frac1t\cdot t=1, $$ whereas \(F(0)=0\). In particular, \(F\) is not continuous at zero and has no finite derivative there. The quotients at zero are \(q_h=0\) for \(h<0\), and \(q_h=h^{-2}\chi_{(0,h)}\) for \(0<h<1\). There cannot be a common integrable dominator \(g\) for the positive quotients: such a bound would imply $$ \int_{(0,h)}g\,dm\ge \int_{(0,h)}h^{-2}\,dm=h^{-2}h=\frac1h $$ for every \(0<h<1\). This contradicts integrability of \(g\), since \(\int_{(0,h)}g\,dm\le\int_{(0,1)}g\,dm<\infty\), while \(1/h\) is unbounded as \(h\downarrow0\). The missing domination is precisely what prevents the criterion from applying.
Common pitfall. The incorrect calculation is “\(\partial_t f(0,x)=0\) for every \(x\), therefore \(F'(0)=\int_0^1 0\,dx=0\).” It silently interchanges a pointwise limit and an integral. In the example, the difference quotient has integral \(1/h\) for positive \(h<1\), not a value tending to zero. The correction is to establish a single integrable bound for all sufficiently small nonzero increments before using Dominated Convergence.
In applications, the practical task is usually to find this bound. Differentiate the integrand with respect to the parameter, bound that derivative uniformly over a parameter interval by an integrable function of \(x\), and apply the scalar mean value theorem to control every difference quotient. The derivative bound must hold on the whole interval between \(t_0\) and \(t_0+h\), not merely at \(t_0\). The resulting criterion works on finite and infinite measure spaces alike; what matters is integrability of the dominator, not finiteness of the underlying measure.
Check Your Understanding
For each question, identify the hypotheses that justify passing the parameter limit through the integral.
- In the exponential example, why does the bound \(ex\) work for every \(|h|<1\), rather than only for one fixed increment?
- Suppose the hypotheses of the difference-quotient criterion hold except that the common bound \(g\) is not integrable. Which step of its proof can no longer invoke Dominated Convergence?
- For \(f(t,x)=e^{-x}\sin(tx)\), verify directly that the pointwise derivative at \(t=0\) is \(xe^{-x}\).
- In the discontinuous-integral example, compute \(\int_0^1 |q_h(x)|\,dx\) for \(0<h<1\). Why does this rule out the conclusion predicted by a false pointwise-limit interchange?
- State the additional derivative bound in the mean-value corollary and explain why it must hold throughout a neighborhood of \(t_0\).