Proof of Differentiation under the Integral Sign

Fix $t\in U$. Being an interior point of $U$, $t$ admits a real $r>0$ such that every real $s$ with $|s-t|<r$ lies in $U$. Call a sequence $(h_k)_{k\in\mathbb{N}}$ of real numbers admissible if $h_k\ne0$ and $t+h_k\in U$ for every $k$ and $(h_k)_{k\in\mathbb{N}}$ has limit $0$; this is the notion used in Sequential Criterion for Differentiability at an Interior Point at the point $t$.

Admissible sequences exist. Indeed, let $(\eta_k)_{k\in\mathbb{N}}$ be a sequence of positive reals with limit $0$, which exists by Existence of a Sequence of Positive Real Numbers with Limit Zero, and set $h_k=\eta_k$ when $\eta_k<r$ and $h_k=r/2$ otherwise. Then $h_k>0$ and $|h_k|<r$, so $t+h_k\in U$; and since $(\eta_k)_{k\in\mathbb{N}}$ has limit $0$ there is $K$ with $\eta_k<r$ for all $k\ge K$, so $h_k=\eta_k$ for $k\ge K$ and $(h_k)_{k\in\mathbb{N}}$ has limit $0$.

Now let $(h_k)_{k\in\mathbb{N}}$ be any admissible sequence and define $u_k:X\to\mathbb{R}$ by

Step 1: each $u_k$ is integrable, with integral the difference quotient of $F$. By hypothesis (i) the functions $x\mapsto f(t+h_k,x)$ and $x\mapsto f(t,x)$ are integrable, so by Linearity and Monotonicity of the Lebesgue Integral their linear combination $u_k$ is integrable, is in particular measurable, and

Step 2: pointwise convergence. Fix $x\in X$. By hypothesis (ii) the function $s\mapsto f(s,x)$ is differentiable at $t$ with derivative $D_1f(t,x)$, and $u_k(x)$ is exactly its difference quotient at $t$ along $h_k$. Claim 1 of Sequential Criterion for Differentiability at an Interior Point therefore gives that $(u_k(x))_{k\in\mathbb{N}}$ has limit $D_1f(t,x)$.

Step 3: domination. Fix $x\in X$ and $k\in\mathbb{N}$. Let $a$ be the smaller and $b$ the larger of the two numbers $t$ and $t+h_k$; since $h_k\ne0$ we have $a<b$, and both lie in $U$. By hypothesis (ii) the function $s\mapsto f(s,x)$ is differentiable at every point of $U$, so Mean Value Theorem on an Open Interval supplies a real $c$ with $a<c<b$ and

If $h_k>0$ then $a=t$ and $b=t+h_k$, so $b-a=h_k$ and the display reads $f(t+h_k,x)-f(t,x)=D_1f(c,x)h_k$. If $h_k<0$ then $a=t+h_k$ and $b=t$, so $b-a=-h_k$ and the display reads $f(t,x)-f(t+h_k,x)=-D_1f(c,x)h_k$. In both cases $u_k(x)=D_1f(c,x)$. Since $a<c<b$ and both $a$ and $b$ lie in the interval $U$, also $c\in U$, so hypothesis (iii) gives $|u_k(x)|\le g(x)$.

Step 4: passage to the limit. By steps 1, 2 and 3 the sequence $(u_k)_{k\in\mathbb{N}}$ consists of measurable functions, converges at every point of $X$ to the function $x\mapsto D_1f(t,x)$, and satisfies $|u_k(x)|\le g(x)$ for every $x$ and every $k$ with $g$ integrable. So Dominated Convergence Theorem applies: the function $x\mapsto D_1f(t,x)$ is measurable and integrable, and

Because admissible sequences exist, this already proves the measurability and integrability asserted for $x\mapsto D_1f(t,x)$.

By step 1 the left-hand side is the difference quotient $\bigl(F(t+h_k)-F(t)\bigr)/h_k$ of $F$ at $t$ along $h_k$. Since the admissible sequence was arbitrary, claim 2 of Sequential Criterion for Differentiability at an Interior Point shows that $F$ is differentiable at $t$ with

Finally $t\in U$ was arbitrary, which gives the conclusion at every point of $U$.