Differentiation under the integral sign

Fix $t_0\in I$ and let $h_n\to 0$ with $h_n\mathrm{\ne }0$ and $t_0+h_n\in I$. Define $q_n(x):=\frac{f(x,t_0+h_n)-f(x,t_0)}{h_n}\mathrm{.}$ For every $x\in X\setminus N_{\ast }$, differentiability in $t$ gives $q_n(x)\to {\mathrm{\partial }}_{t}f(x,t_0)$. Give the derivative value zero on $N_{\ast }$; this measurable modification differs from the stated zero extension only on a null set, so it has the same integral.

For each $n$, hypothesis 1 makes $x\mapsto f(x,t_0+h_n)$ and $x\mapsto f(x,t_0)$ integrable and therefore measurable, so $q_n$ is measurable. Fix $x\in X\setminus N_{\ast }$ and $n$. If $q_n(x)=0$, then $\mathrm{\mid }{q}_n(x)\mathrm{\mid }\le g(x)$ is immediate. Otherwise put $\alpha :=\overline{q_n(x)}\mathrm{/}\mathrm{\mid }{q}_n(x)\mathrm{\mid },$ so $\mathrm{\mid }\alpha \mathrm{\mid }=1$ and $\mathrm{\mid }{q}_n(x)\mathrm{\mid }=\mathrm{Re}\text{\hspace{0.17em}⁣}\left(\alpha \frac{f(x,t_0+h_n)-f(x,t_0)}{h_n}\right)\mathrm{.}$ Apply [L2] to the real-valued function $\tau \mapsto \mathrm{Re}(\alpha f(x,\tau ))$ on the segment joining $t_0$ to $t_0+h_n$. For some interior point $\xi$ of that segment, $\mathrm{\mid }{q}_n(x)\mathrm{\mid }=\mathrm{Re}(\alpha {\mathrm{\partial }}_{t}f(x,\xi ))\le \mathrm{\mid }{\mathrm{\partial }}_{t}f(x,\xi )\mathrm{\mid }\le g(x)\mathrm{.}$ Hypothesis 3 and the zero extension make the limit measurable. Therefore [L1] applies to $(q_n)$.

By [L1], ${\lim }_n\int q_n(x)d\mu (x)=L:=\int {\mathrm{\partial }}_{t}f(x,t_0)d\mu (x)\mathrm{.}$ Linearity of the integral gives $\int q_n(x)d\mu (x)=\frac{F(t_0+h_n)-F(t_0)}{h_n}\mathrm{.}$ This holds for every supplied sequence of admissible nonzero increments.

The same mean-value estimate as in step 2.1, with any $s,t\in I$ in place of $t_0,t_0+h_n$, gives $\mathrm{\mid }f(x,t)-f(x,s)\mathrm{\mid }\le \mathrm{\mid }t-s\mathrm{\mid }g(x)$ outside $N_{\ast }$. Integrating yields $\mathrm{\mid }F(t)-F(s)\mathrm{\mid }\le \mathrm{\mid }t-s\mathrm{\mid }\int gd\mu$, so $F$ is continuous. Consequently $Q(h)=(F(t_0+h)-F(t_0))\mathrm{/}h$ is continuous on the punctured interval of admissible increments. If $Q(h)$ failed to tend to $L$ as $h\to 0$, there would be an $\epsilon >0$ and, for every $n\ge 1$, a nonzero admissible $h$ with $\mathrm{\mid }h\mathrm{\mid }<1\mathrm{/}n$ and $\mathrm{\mid }Q(h)-L\mathrm{\mid }\ge \epsilon$. Continuity of $Q$ and [L4] give a rational admissible $r$ with $\mathrm{\mid }r\mathrm{\mid }<1\mathrm{/}n$ and $\mathrm{\mid }Q(r)-L\mathrm{\mid }>\epsilon \mathrm{/}2$. Fix an enumeration of $\mathbb{Q}$ and take the first such rational $r_n$ for each $n$; this is a definable selection from a countable set and uses no Countable Choice. Then $r_n\to 0$, contradicting step 3.1. Thus $Q(h)\to L$, which is precisely $F^{\mathrm{\prime }}(t_0)=L$. Since $t_0$ was arbitrary, the theorem follows.