The Local Differentiation Theorem
Differentiating an integral is not a consequence of pointwise differentiability alone. The key issue is whether difference quotients can be controlled by one integrable function as the parameter varies. When such a bound is available, the Dominated Convergence Theorem gives a rigorous passage from the pointwise derivative to the derivative of the integral.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Let \(I\subseteq\mathbb{R}\) be an open interval, let \(t_0\in I\), and let \(f:I\times X\to\mathbb{R}\). Assume that \(x\mapsto f(t,x)\) is \(\mathcal{M}\)-measurable for every \(t\in I\). For \(t\in I\), write \(F(t)=\int_X f(t,x)\,d\mu(x)\) whenever this is a finite Lebesgue integral.
The value assigned to \(D\) on \(N\) is part of the theorem’s formulation: it gives a measurable version of the derivative. On a noncomplete measure space, an arbitrary assignment on a measurable null set need not be measurable. The integral is independent of the choice among measurable versions that agree \(\mu\)-a.e.
Proof by Difference Quotients
Proof. Since \(g\) is integrable, it is finite \(\mu\)-a.e. Enlarge \(N\), if necessary, by the measurable null set \(\{x:g(x)=\infty\}\). This preserves the hypotheses, and now \(g(x)<\infty\) for every \(x\notin N\).
First fix \(t\in J\). For \(x\notin N\), the one-variable Mean Value Theorem applied on the interval between \(t_0\) and \(t\) gives \(\lvert f(t,x)-f(t_0,x)\rvert\le \lvert t-t_0\rvert g(x)\). Consequently, \(\lvert f(t,x)\rvert\le \lvert f(t_0,x)\rvert+\lvert t-t_0\rvert g(x)\) outside \(N\). The right side is integrable. The section \(f(t,\cdot)\) is measurable by assumption, so this a.e. bound proves it is integrable. In particular, \(F(t)\) is a finite real number for each \(t\in J\).
For \(h\ne0\) with \(t_0+h\in J\), define the measurable difference quotient \(q_h(x)=(f(t_0+h,x)-f(t_0,x))/h\). For \(x\notin N\), the Mean Value Theorem and the derivative bound give \(\lvert q_h(x)\rvert\le g(x)\). Also, \(q_h(x)\to\partial_t f(t_0,x)\) as \(h\to0\) for every \(x\notin N\).
We must ensure the limiting function is measurable, not merely defined almost everywhere. Choose any sequence \(h_n\ne0\) with \(h_n\to0\) and \(t_0+h_n\in J\). The functions \(r_n=\chi_{N^c}q_{h_n}\) are measurable. Their pointwise limit is \(D\): outside \(N\), this follows from differentiability, and on \(N\), every \(r_n\) is zero. Thus \(D\) is measurable. The bound \(\lvert D\rvert\le g\) holds outside \(N\), so \(D\) is integrable.
Now take any sequence of nonzero real numbers \(h_n\to0\) with \(t_0+h_n\in J\). The difference quotients \(q_{h_n}\) are measurable, converge to \(D\) \(\mu\)-a.e., and satisfy \(\lvert q_{h_n}\rvert\le g\) \(\mu\)-a.e., where \(g\) is integrable. The Dominated Convergence Theorem therefore gives \(\int_X q_{h_n}\,d\mu\to\int_X D\,d\mu\). By linearity of the Lebesgue integral, \(\int_X q_{h_n}\,d\mu=(F(t_0+h_n)-F(t_0))/h_n\). Since this limit holds for every such sequence, the sequential criterion for a real limit proves that \(F'(t_0)=\int_X D\,d\mu\). Both sides are finite because \(F(t_0+h)\) is finite for \(t_0+h\in J\) and \(D\) is integrable. \(\square\)
Applications and a Failed Hypothesis
Worked Example: A Smooth Integrand on a Finite Interval
Take \(X=[0,1]\), \(\mathcal{M}\) the Lebesgue sigma-algebra, \(\mu=m\), and \(f(t,x)=e^{tx}\) for \(t\in(-1,1)\). Find \(F'(0)\), where \(F(t)=\int_0^1 e^{tx}\,dm(x)\).
For each fixed \(t\), the section is measurable, and \(f(0,x)=1\) is integrable. For every \(x\in[0,1]\), \(\partial_t f(t,x)=xe^{tx}\). If \(\lvert t\rvert<1\), then \(0\le xe^{tx}\le e\): indeed \(x\le1\) and \(e^{tx}\le e^{\lvert t\rvert x}\le e\). Thus \(g(x)=e\) on \([0,1]\) is an integrable bound, with \(\int_0^1g\,dm=e\). The theorem applies with \(N=\varnothing\), and gives \(F'(0)=\int_0^1 xe^{0x}\,dm(x)=\int_0^1x\,dm(x)=\tfrac12\).
Check: Direct integration gives \(F(t)=(e^t-1)/t\) for \(t\ne0\), while \(F(0)=1\). The power series \(e^t=1+t+t^2/2+o(t^2)\) yields \((F(t)-F(0))/t=((e^t-1)/t-1)/t\to\tfrac12\), confirming the derivative.
Worked Example: Differentiating a Series Using Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure \(\mu\), and set \(f(t,n)=(1+t)2^{-n}\) for \(t\in(-\tfrac12,\tfrac12)\). Compute the derivative of \(F(t)=\int_X f(t,n)\,d\mu(n)\) at \(t=0\).
Every function on this measurable space is measurable. The section at \(t=0\) is integrable because \(\sum_{n=1}^{\infty}2^{-n}=1\). For each \(n\), \(\partial_t f(t,n)=2^{-n}\), so the integrable dominating function is \(g(n)=2^{-n}\), whose counting-measure integral is \(1\). The theorem gives \(F'(0)=\sum_{n=1}^{\infty}2^{-n}=1\).
Independently, linearity and the geometric series give \(F(t)=(1+t)\sum_{n=1}^{\infty}2^{-n}=1+t\), hence \(F'(0)=1\). The integrability and domination hypotheses have both been checked; this is not a formal term-by-term differentiation without a convergence argument.
Worked Example: Pointwise Derivatives Do Not Suffice
On \(X=[0,\infty)\) with Lebesgue measure, define \(\phi(u)=(u-1)^2(2-u)^2\) for \(1<u<2\), and \(\phi(u)=0\) otherwise. For \(t\in(-1,1)\), set \(f(t,x)=t^{-1}\phi(x/t)\) when \(t>0\), and \(f(t,x)=0\) when \(t\le0\). Show why differentiating the integral at \(t=0\) by merely integrating the pointwise derivative fails.
The function \(\phi\) and its first derivative vanish at the endpoints \(1\) and \(2\), so its extension by zero is continuously differentiable. For each fixed \(x>0\), \(f(t,x)=0\) whenever \(0<t<x/2\), since then \(x/t>2\). For \(t\le0\) it is also zero. Thus \(t\mapsto f(t,x)\) is differentiable at zero with derivative zero. At \(x=0\), it is identically zero, so its derivative is again zero.
For \(t>0\), the section is supported where \(t<x<2t\), and a change of variables \(u=x/t\) computes its integral exactly: \(\int_0^\infty f(t,x)\,dm(x)=\int_1^2\phi(u)\,du\). Writing \(v=u-1\), this last integral is \(\int_0^1v^2(1-v)^2\,dv=\int_0^1(v^2-2v^3+v^4)\,dv=\tfrac13-\tfrac12+\tfrac15=\tfrac1{30}\). For \(t\le0\), the integral is zero. Hence \(F(0)=0\) but \(F(t)=1/30\) for \(t>0\), and \(F\) is not continuous at zero. The integral of the pointwise derivative is zero, but \(F'(0)\) does not exist.
Check: each section \(f(t,\cdot)\) is integrable: it is zero for \(t\le0\), and for \(t>0\) its integral is the finite value \(1/30\). The missing hypothesis is a single integrable bound for \(\lvert\partial_t f(t,x)\rvert\) in a neighborhood of zero. Pointwise differentiability alone does not provide it.
Using the Theorem Correctly
The most common error is to write \(\frac{d}{dt}\int_X f(t,x)\,d\mu(x)=\int_X\partial_t f(t,x)\,d\mu(x)\) as if pointwise differentiability justified the equality. The example above is a concrete counterexample: the right side at zero is \(0\), while the integral is discontinuous there. The proof requires all of the following local checks:
- The sections \(f(t,\cdot)\) are measurable, and the section at \(t_0\) is integrable.
- Outside one measurable null set, the parameter functions are differentiable throughout a neighborhood of \(t_0\).
- Their derivatives on that neighborhood are bounded in absolute value by one integrable function independent of the parameter.
- The derivative used in the final integral is a measurable version, such as the version defined to be zero on the exceptional null set.
The bound is what controls the entire family of difference quotients, not just the pointwise derivative at \(t_0\). It also proves that nearby sections are integrable, so the parameter integral is well-defined before its derivative is considered. The argument establishes differentiability at one interior parameter value; it does not by itself assert repeated differentiability or handle moving integration domains.
Check Your Understanding
For each question, identify the measure space and the hypotheses that justify—or prevent—the differentiation step.
- For \(f(t,x)=e^{tx}\) on \([0,1]\), find an integrable bound for \(\lvert\partial_t f(t,x)\rvert\) when \(\lvert t\rvert<1\), and state \(F'(0)\).
- In the theorem, why must the derivative on the exceptional null set be chosen as a measurable version rather than assigned arbitrary values?
- For counting measure on \(\mathbb{N}\), apply the theorem to \(f(t,n)=(1+t)3^{-n}\) near \(t=0\) and compute \(F'(0)\).
- In the moving-support example, compute \(F(t)\) for \(t>0\) and explain why its pointwise derivative at zero does not determine \(F'(0)\).
- Which part of the theorem’s proof uses the Mean Value Theorem, and which part uses the Dominated Convergence Theorem?