Differentiation under the integral sign over a compact measurable set and over a compact $C^1$ surface

The function $F(t):={\int }_{K}f(x,t)d{\lambda }_n(x)$ is defined and finite for every $t\in [a,b]$: for fixed $t$ the map $x\mapsto f(x,t)$ is continuous on $K$, hence bounded and Borel measurable, and ${\lambda }_n(K)<\mathrm{\infty }$, so it is integrable by [F2].

Fix $s,t\in [a,b]$ with $s\mathrm{\ne }t$ and put $g(x):=\frac{f(x,s)-f(x,t)}{s-t}$ for $x\in K$; then $g$ is continuous on $K$, hence bounded, Borel measurable and integrable by [F2]. For every $x\in K$ the mean value theorem [F1] applied to the one-variable function $u\mapsto f(x,u)$ on the closed interval with endpoints $t$ and $s$ supplies a point $\tau$ strictly between $t$ and $s$ with $f(x,s)-f(x,t)={\mathrm{\partial }}_{t}f(x,\tau )(s-t)$, that is, $g(x)={\mathrm{\partial }}_{t}f(x,\tau )$; no point $\tau$ is selected as a function of $x$ anywhere below, only the pointwise inequality $\mathrm{\mid }g(x)-{\mathrm{\partial }}_{t}f(x,t)\mathrm{\mid }<\eta$ that such an existential $\tau$ yields, together with the separately defined continuous function $g$. By linearity of the integral [F3] applied to the integrable functions $f(\cdot ,s)$ and $f(\cdot ,t)$ of step 1.1, $F(s)-F(t)={\int }_K(f(x,s)-f(x,t))d{\lambda }_n(x)=(s-t){\int }_{K}gd{\lambda }_n(x)$, so the difference quotient is ${\int }_{K}gd{\lambda }_n$.

Given $\epsilon >0$, put $\eta :=\epsilon \mathrm{/}(1+{\lambda }_n(K))>0$; the function ${\mathrm{\partial }}_{t}f$ is uniformly continuous on the compact set $K\times [a,b]$ by [F4], so there is $\delta >0$ such that $\mid {\mathrm{\partial }}_{t}f(x,u)-{\mathrm{\partial }}_{t}f(x,v)\mid <\eta$ whenever $(x,u),(x,v)\in K\times [a,b]$ with $\mid u-v\mid <\delta$. Fix $s,t\in [a,b]$ with $0<\mid s-t\mid <\delta$ and let $g$ be the continuous function of step 2.1. For every $x\in K$ the mean value point $\tau$ supplied by [F1] lies strictly between $t$ and $s$, so $\mid \tau -t\mid <\mid s-t\mid <\delta$ and therefore $\mid g(x)-{\mathrm{\partial }}_{t}f(x,t)\mid =\mid {\mathrm{\partial }}_{t}f(x,\tau )-{\mathrm{\partial }}_{t}f(x,t)\mid <\eta$; the function $h:=g-{\mathrm{\partial }}_{t}f(\cdot ,t)$ is continuous on $K$, hence bounded and Borel measurable, so by [F2] and [F3] $\mid \frac{F(s)-F(t)}{s-t}-{\int }_K{\mathrm{\partial }}_{t}f(x,t)d{\lambda }_n(x)\mid =\mid {\int }_{K}hd{\lambda }_n\mid \le {\int }_K\mid h\mid d{\lambda }_n\le {\int }_K\eta d{\lambda }_n=\eta {\lambda }_n(K)=\frac{\epsilon {\lambda }_n(K)}{1+{\lambda }_n(K)}<\epsilon$; the case ${\lambda }_n(K)=0$ is covered by the same chain, both sides vanishing. No mean-value point is selected as a function of $x$ in this estimate; ${\mathrm{A}\mathrm{C}}_{\omega }$ has already supplied the Lebesgue measure-space premise [A1].

Step 3.1 shows that for every $t\in [a,b]$ and every sequence $h_k\to 0$ with $h_k\mathrm{\ne }0$ and $t+h_k\in [a,b]$ the difference quotients converge to ${\int }_K{\mathrm{\partial }}_{t}f(x,t)d{\lambda }_n(x)$: for interior $t$ this is differentiability with the stated derivative, and at $t=a$ (respectively $t=b$), where only $h_k>0$ (respectively $h_k<0$) is admissible, it is the relative derivative there; because $N$ is open, ${\mathrm{\partial }}_{t}f(x,t)$ is the genuine two-sided partial derivative, so the relative endpoint derivative exists and equals the displayed integral. This proves the set form.

Surface form: let $G(t):={\int }_{S}f(\sigma ,t)dS(\sigma )$, a finite real number for each $t$ by [F5]. Fix $s,t\in [a,b]$ with $s\mathrm{\ne }t$ and put $g(\sigma ):=\frac{f(\sigma ,s)-f(\sigma ,t)}{s-t}$; $g$ is continuous on the compact $S$, hence $\sigma \mapsto g(\sigma )-{\mathrm{\partial }}_{t}f(\sigma ,t)$ is continuous, bounded and integrable against $dS$, and by linearity of the surface integral [F5] on the integrable functions $G(s)-G(t)={\int }_S(f(\sigma ,s)-f(\sigma ,t))dS(\sigma )=(s-t){\int }_{S}gdS(\sigma )$. For every $\sigma \in S$ the mean value theorem [F1] supplies $\tau$ strictly between $t$ and $s$ with $g(\sigma )={\mathrm{\partial }}_{t}f(\sigma ,\tau )$, and again no $\tau$ is selected as a function; uniform continuity of ${\mathrm{\partial }}_{t}f$ on the compact set $S\times [a,b]$ by [F4], the finiteness of the surface measure $S(S)$ and the monotonicity and triangle inequality for the surface integral [F5] then give, for $0<\mid s-t\mid <\delta$ with the $\delta$ of step 3.1 applied to $\eta :=\epsilon \mathrm{/}(1+S(S))$, the pointwise bound $\mid g(\sigma )-{\mathrm{\partial }}_{t}f(\sigma ,t)\mid <\eta$ and hence $\mid [G(s)-G(t)]\mathrm{/}(s-t)-{\int }_S{\mathrm{\partial }}_{t}f(\sigma ,t)dS(\sigma )\mid =\mid {\int }_S(g-{\mathrm{\partial }}_{t}f(\cdot ,t))dS\mid \le \eta S(S)<\epsilon$, exactly as in steps 2.1 and 3.1 with $S$ in place of $K$ and $dS$ in place of $d{\lambda }_n$. Passing to the limit as in step 4.1 gives the surface form, including the relative endpoint derivatives.