TITLE: Passing derivative into integral without using dominated convergence theorem
QUESTION [1 upvotes]: Let $f(x,y)$ be continuous on $[a,b] \times [c,d]$, $\frac{\partial f}{\partial x}(x,y)$ exists and is continuous at every point of the domain. $\Phi(x)=\int_c^d f(x,y) dy$ is differentiable with $\Phi'(x)=\int_c^d \frac{\partial f}{\partial x}(x,y) dy$.
This is a well known result, but as far as I know its proof uses the dominated convergence theorem in one way or another. Is there a proof that uses more basic techniques, by that I mean only those results accessible to students who have just learnt Riemann integration and convergence of sequence of functions?

REPLY [1 votes]: Fix $x \in [a,b]$. If $x+h \in [a,b]$, and $h \neq 0$, then $$\frac{1}{h}(\Phi(x+h) - \Phi(x)) = \int_c^d \frac{f(x+h,y)-f(x,y)}{h}dy.$$ For each $c \leqq y\leqq d$, we can apply the mean value theorem to the map $t \mapsto f(t,y)$ to obtain $0<\theta_y<1$ such that $$f(x+h,y) - f(x,y) = h\frac{\partial f }{\partial x}(x+\theta_yh,y).$$ Let $\epsilon > 0$ be given. By the uniform continuity of $\partial f/\partial x$ on $[a,b]\times[c,d]$, there exists a $\delta>0$ for which $$\left|\frac{\partial f}{\partial x}(s^\prime,t^\prime) - \frac{\partial f}{\partial x}(s,t)\right|<\epsilon,$$ if $|s^\prime-s|<\delta$ and $|t^\prime-t|<\delta$. Thus, if $0<|h| < \delta$ we have $$\left| \frac{1}{h}(\Phi(x+h) - \Phi(x)) - \int_c^d \frac{\partial f}{\partial x}(x,y)dy\right| \leqq \int_c^d \left|\frac{\partial f}{\partial x}(x+\theta_y h,y) - \frac{\partial f}{\partial x}(x,y)\right|dy \leqq \epsilon (d-c).$$
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