Definite Integral of Partial Derivative/Proof 2

Proof
Define $\displaystyle G \left({x}\right) = \int_a^b f \left({x, y}\right) \rd y$.

The continuity of $f$ ensures that $G$ exists.

Then by linearity of the integral:
 * $\displaystyle \frac {\Delta G}{\Delta x} = \frac{G \left({x + \Delta x}\right) - G \left({x}\right)} {\Delta x} = \int_a^b \frac {f \left({x + \Delta x, y}\right) - f \left({x, y}\right)} {\Delta x} \rd y$

We want to find the limit of this quantity as $\Delta x$ approaches zero.

For each $y \in \left[{a \,.\,.\, b}\right]$, we can consider $f_y \left({x}\right) = f \left({x, y}\right)$ as a separate function of the single variable $x$, with $\displaystyle \frac {\rd f_y} {\rd x} = \frac {\partial f} {\partial x}$.

Thus by the Mean Value Theorem, there is a number $c_y \in \left({x, x + \Delta x}\right)$ such that $\displaystyle f_y \left({x + \Delta x}\right) - f_y \left({x}\right) = \frac {\rd f_y} {\rd x} \left({c_y}\right) \Delta x$.

That is:
 * $\displaystyle f \left({x + \Delta x, y}\right) - f \left({x, y}\right) = \frac {\partial f} {\partial x} \left({c_y, y}\right) \Delta x$

Therefore:
 * $\displaystyle \frac {\Delta G} {\Delta x} = \int_a^b \frac {\partial f} {\partial x} \left({c_y, y}\right) \rd y$

Now, pick any $\epsilon > 0$ and set $\epsilon_0 = \dfrac {\epsilon} {b - a}$.

Since $\dfrac {\partial f} {\partial x}$ is continuous on the compact set $D$, it is uniformly continuous on $D$.

Hence for each $x$ and $y$:
 * $\displaystyle \left\vert{\frac {\partial f} {\partial x} \left({x + h, y}\right) - \frac {\partial f} {\partial x} \left({x, y}\right)}\right\vert < \epsilon_0$

whenever $h$ is sufficiently small.

And since $x < c_y < x + \Delta x\ $, it follows that for sufficiently small $\Delta x\ $ that:
 * $\displaystyle \left\vert{\frac {\partial f} {\partial x} \left({c_y, y}\right) - \frac {\partial f} {\partial x} \left({x, y}\right)}\right\vert < \epsilon_0$

regardless of our choice of $y$.

So we can say:

But since $\epsilon$ was arbitrary, it follows that $\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} = \int_a^b \frac {\partial f} {\partial x} \left({x, y}\right) \rd y$ and the theorem is proved.