Proof of Differentiation under the Integral Sign

Throughout, integrable functions are measurable (The Borel Sets of ℝ Form a σ-Algebra, and Integrable Functions Are Measurable Real-Valued Maps (iii)), and continuity of a real function at a point of an interval is used in its $\varepsilon$–$\delta$ form (Continuity of Real Functions on Subsets of ℝ in ε-δ Form (i)).

We begin with (i). Fix $t\in I$. For $s\in I$ with $s\neq t$ define the difference quotient $Q_{s}:X\to\mathbb{R}$,

$$Q_{s}(x)=\frac{f(s,x)-f(t,x)}{s-t}.$$

As a linear combination of $x\mapsto f(s,x)$ and $x\mapsto f(t,x)$, both in $\mathcal{L}^{1}(\mu)$, $Q_{s}$ lies in $\mathcal{L}^{1}(\mu)$, and by linearity of the integral

$$\int_{X}Q_{s}\,d\mu=\frac{F(s)-F(t)}{s-t}.$$

For each $x\in X$ the function $s'\mapsto f(s',x)$ is differentiable, hence continuous, on $I$ (The Increment Form of the Derivative, and Differentiability Implies Continuity (ii)). By the mean value theorem, Mean Value Theorem on a Closed Real Interval and on Any Interval (ii), applied to this function on $I$ with the points $\min(s,t)<\max(s,t)$ of $I$ (the quotient $Q_{s}(x)$ is unchanged when $s$ and $t$ are exchanged), there is $\xi\in I$ strictly between $s$ and $t$ with $Q_{s}(x)=\partial_{t}f(\xi,x)$, the derivative of $s'\mapsto f(s',x)$ at $\xi$. Since $\xi\in I$, the hypothesis gives

$$|Q_{s}(x)|\le g(x)\qquad\text{for all }x\in X\text{ and }s\in I\setminus\{t\}.$$

Next we show that $x\mapsto\partial_{t}f(t,x)$ is integrable. Let $s_{j}=t+(q-t)/(j+1)$ for $j\in\mathbb{N}$; then $s_{j}\in I$, $s_{j}\neq t$ and $s_{j}\to t$. By the definition of the derivative, since $s_{j}\to t$ with $s_{j}\neq t$, $Q_{s_{j}}(x)\to\partial_{t}f(t,x)$ for every $x\in X$. Thus $x\mapsto\partial_{t}f(t,x)$ is a pointwise limit of measurable functions, hence measurable, and the sequence $(Q_{s_{j}})$ is dominated by the integrable function $g$. The dominated convergence theorem, Dominated Convergence Theorem (i) with dominating function $g$, shows that $x\mapsto\partial_{t}f(t,x)$ lies in $\mathcal{L}^{1}(\mu)$, and we put

$$L=\int_{X}\partial_{t}f(t,x)\,d\mu(x).$$

It remains to show that $F$ is differentiable at $t$ with $F'(t)=L$. The set $\mathbb{Q}$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ix) and nonempty, since $0\in\mathbb{Q}$; so, by Countable Set (i), there is a surjective map $\mathbb{N}\to\mathbb{Q}$, $k\mapsto q_{k}$, and we fix one such enumeration of $\mathbb{Q}$. For $j\in\mathbb{N}$ let

$$P_{j}=\{s\in I:0<|s-t|<1/j\},$$

and for $j,k\in\mathbb{N}$ define $\varphi_{j,k}:X\to\mathbb{R}$ by $\varphi_{j,k}(x)=|Q_{q_{k}}(x)-\partial_{t}f(t,x)|$ if $q_{k}\in P_{j}$, and $\varphi_{j,k}(x)=0$ otherwise. Each $\varphi_{j,k}$ is measurable: in the first case $Q_{q_{k}}$ and $x\mapsto\partial_{t}f(t,x)$ lie in $\mathcal{L}^{1}(\mu)$, hence are measurable, and so is $\varphi_{j,k}$ by Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i); in the second case $\varphi_{j,k}$ is constant, hence measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii). For $s\in I\setminus\{t\}$ and $x\in X$, the bound on difference quotients and the hypothesis give

$$|Q_{s}(x)-\partial_{t}f(t,x)|\le|Q_{s}(x)|+|\partial_{t}f(t,x)|\le2g(x),$$

so $0\le\varphi_{j,k}(x)\le2g(x)$ for all $j,k\in\mathbb{N}$ and $x\in X$. Hence, for each $j\in\mathbb{N}$, the supremum

$$h_{j}(x)=\sup_{k\in\mathbb{N}}\varphi_{j,k}(x)$$

exists in $\mathbb{R}$, the function $h_{j}:X\to\mathbb{R}$ is measurable by Suprema, Pointwise Limits and the Convergence Set of Measurable Real-Valued Maps (i), and $0\le h_{j}\le2g$. In words, $h_{j}(x)$ is the supremum of $|Q_{r}(x)-\partial_{t}f(t,x)|$ over the rational $r\in P_{j}$, with $0$ added.

The function $h_{j}$ also controls the difference quotients at the real points of $P_{j}$:

$$|Q_{s}(x)-\partial_{t}f(t,x)|\le h_{j}(x)\qquad\text{for all }j\in\mathbb{N},\ s\in P_{j}\text{ and }x\in X .$$

To see this, fix $j$, $s\in P_{j}$ and $x$, and put $d=|s-t|$, so that $0<d<1/j$. Let $\varepsilon>0$. First, since $s'\mapsto f(s',x)$ is continuous on $I$, there is $\eta>0$ such that every $r\in I$ with $|r-s|<\eta$ satisfies $|f(r,x)-f(s,x)|<\varepsilon d/4$. Next put

$$\rho=\min\Bigl\{\eta,\ \frac{d}{2},\ \frac{\varepsilon d}{4(|Q_{s}(x)|+1)},\ s-p,\ q-s,\ \frac{1}{j}-d\Bigr\}>0 ,$$

and, by the density of $\mathbb{Q}$ in $\mathbb{R}$ (The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (x)), take $r\in\mathbb{Q}$ with $s-\rho<r<s+\rho$; then $r=q_{k}$ for some $k\in\mathbb{N}$. Since $|r-s|<\rho$, we have $p<r<q$, $|r-t|\ge d-|r-s|>d/2>0$ and $|r-t|\le d+|r-s|<1/j$, so $r\in P_{j}$ and $\varphi_{j,k}(x)=|Q_{r}(x)-\partial_{t}f(t,x)|$. Multiplying out the definitions of $Q_{r}$ and $Q_{s}$ gives

$$(r-t)\bigl(Q_{r}(x)-Q_{s}(x)\bigr)=f(r,x)-f(s,x)-(r-s)\,Q_{s}(x),$$

By the choice of $\eta$, $|f(r,x)-f(s,x)|<\varepsilon d/4$, and by the third term of $\rho$, $|r-s|\,|Q_{s}(x)|\le\varepsilon d/4$. Therefore

$$|Q_{r}(x)-Q_{s}(x)|\le\frac{|f(r,x)-f(s,x)|+|r-s|\,|Q_{s}(x)|}{|r-t|}<\frac{\varepsilon d/4+\varepsilon d/4}{d/2}=\varepsilon .$$

Hence $|Q_{s}(x)-\partial_{t}f(t,x)|\le|Q_{r}(x)-\partial_{t}f(t,x)|+\varepsilon\le h_{j}(x)+\varepsilon$, and as $\varepsilon>0$ was arbitrary, the bound follows.

Moreover, $h_{j}(x)\to0$ for every $x\in X$. Fix $x$ and let $\varepsilon>0$. First, by the definition of the derivative of $s'\mapsto f(s',x)$ at $t$, there is $\delta>0$ such that $|Q_{s}(x)-\partial_{t}f(t,x)|<\varepsilon/2$ for every $s\in I$ with $0<|s-t|<\delta$. Then, by the Archimedean property (The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (ix)), there is $N\in\mathbb{N}$ with $1/\delta<N$. For $j\ge N$ we have $1/j\le1/N<\delta$, so every $\varphi_{j,k}(x)$ is $0$ or a value $|Q_{s}(x)-\partial_{t}f(t,x)|$ with $0<|s-t|<\delta$, hence less than $\varepsilon/2$, and $0\le h_{j}(x)\le\varepsilon/2<\varepsilon$.

The functions $2g$ and $0$ lie in $\mathcal{L}^{1}(\mu)$, with $\int_{X}0\,d\mu=0$, by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), and $|h_{j}|\le2g$ for every $j$. The dominated convergence theorem, Dominated Convergence Theorem (i) and Dominated Convergence Theorem (iii) with dominating function $2g$ and limit function $0$, shows that every $h_{j}$ lies in $\mathcal{L}^{1}(\mu)$ and that

$$\int_{X}h_{j}\,d\mu\longrightarrow0 .$$

Now let $\varepsilon>0$. First choose $j\in\mathbb{N}$ with $\int_{X}h_{j}\,d\mu<\varepsilon$, and then put $\delta=1/j$. Let $s\in I$ with $0<|s-t|<\delta$, that is, $s\in P_{j}$. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), the function $x\mapsto Q_{s}(x)-\partial_{t}f(t,x)$ lies in $\mathcal{L}^{1}(\mu)$ and has integral $(F(s)-F(t))/(s-t)-L$. So, by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x), and then by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix) with the bound by $h_{j}$ above,

$$\Bigl|\frac{F(s)-F(t)}{s-t}-L\Bigr|\le\int_{X}\bigl|Q_{s}(x)-\partial_{t}f(t,x)\bigr|\,d\mu(x)\le\int_{X}h_{j}\,d\mu<\varepsilon .$$

Thus $L$ is a derivative of $F$ at $t$ in the sense of Derivative of a Real Function on an Interval (i), so $F$ is differentiable at $t$ with $F'(t)=L$ (Derivative of a Real Function on an Interval (ii)). As $t\in I$ was arbitrary, this proves (i).

We deduce (ii) by restricting to a line. Fix $u\in U$ and, by Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (xi), choose $\rho>0$ such that $u+se_{i}\in U$ for every real $s$ with $|s|<\rho$, that is, for $s\in(-\rho,\rho)$. Define $\tilde f:(-\rho,\rho)\times X\to\mathbb{R}$ by $\tilde f(s,x)=f(u+se_{i},x)$. For every $s$, the function $x\mapsto\tilde f(s,x)$ lies in $\mathcal{L}^{1}(\mu)$. For $x\in X$ and $s,s+h\in(-\rho,\rho)$ we have $\tilde f(s+h,x)=f((u+se_{i})+he_{i},x)$ by Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (iii). Hence, by Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (i), for every $x\in X$ the function $s\mapsto\tilde f(s,x)$ is differentiable on $(-\rho,\rho)$, with derivative $\partial_{i}f(u+se_{i},x)$ at $s$, and this derivative is bounded by $g(x)$. So $\tilde f$ satisfies the hypotheses of (i) on the interval $(-\rho,\rho)$, applied at the point $0$ (where $u+0e_{i}=u$, by Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (i)), which shows that $x\mapsto\partial_{i}f(u,x)$ lies in $\mathcal{L}^{1}(\mu)$ and that $s\mapsto F(u+se_{i})$ is differentiable at $0$ with derivative

$$\int_{X}\partial_{i}f(u,x)\,d\mu(x).$$

By Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (i) and Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (ii), this number is $\partial_{i}F(u)$.