Passing the derivative inside the integral Let $f(x,y)$, with $0\le x,y\le 1$, be integrable in $y$ for each $x$. Further suppose $\partial f(x,y)/\partial x$ is a bounded function of $(x,y)$. Show that $\partial f(x,y)/\partial x$ is a measurable function of $y$ for each $x$ and
$$\frac{d}{dx}\int_0^1{f(x,y)\,dy} = \int_0^1{\frac{\partial}{\partial x}f(x,y)\,dy}\,.$$
I wish I had some work to really show. At this point all I've done is written down each side of the desired equation in terms of limits, and I know ultimately this problem is actually about passing limits inside integrals... But we've done almost no examples of this sort of thing (we literally just finished the limiting theorems today: Fatou, MCT, DCT, etc), and I'm not sure where to begin.
 A: Write $\frac{d}{dx}\int_0^1 f(x,y)dy$ as $$\lim_{h\to 0}\int_0^1 \frac{f(x+h,y)-f(x,y)}{h}dy.$$(You can do that because integration is linear.) If $$|\frac{\partial f}{\partial x}|\le M,$$ then by mean valued theorem
$$|\frac{f(x+h,y)-f(x,y)}{h}|\le M.$$
Now apply Bounded Convergence Theorem (a weaker form of Lebesgue convergence theorem in this case.)
A: You can write the limits of difference quotients as limits of sequences of difference quotients.  The measurability of the partial derivative follows from the fact that a pointwise limit of a sequence of measurable functions is measurable.  The left-hand side of the equation you want to prove is a limit of integrals of difference quotients, and the right-hand side is an integral of limits of difference quotients, so you want to justify passing a limit through an integral.  You can do this because the sequence of difference quotients is uniformly bounded, by the mean value theorem and the hypothesis that $\frac{\partial f}{\partial x}$ is bounded. 
