# 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.

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Seen this? – J. M. Nov 24 '10 at 5:22
Just to clarify, you are assuming that $\frac{\partial f}{\partial x}$ exists everywhere, not just a.e.? – Jonas Meyer Nov 24 '10 at 5:31
@Jonas: I transcribed the problem almost verbatim. The book does not specify – Bey Nov 24 '10 at 15:29
I assume it means everywhere, because otherwise extra hypotheses would be needed. I gave an answer with this assumption (hence the application of the mean value theorem). TCL has done the same. – Jonas Meyer Nov 24 '10 at 18:11

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 $$|\frac{f(x+h,y)-f(x,y)}{h}|\le M.$$
By MVT $f(x+h,y)-f(x,y)=h\frac{\partial f}{\partial x}(\xi,y)$ for some $\xi$ between $x+h$ and $x$. So $\left|\frac{f(x+h,y)-f(x,y)}{h}\right|=\left|\frac{\partial f}{\partial x}(\xi,y)\right|\le M.$ – TCL Nov 26 '10 at 4:30
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.