# Using dominated convergence theorem to asset Leibniz rule when $\partial{f}/\partial{x}(x,y)$ is bounded and $f$ is integrable

I was thinking about the following problem

Let $f(x,y)$, $0 \leq x, y \leq 1$, be a function such that for each $x$, $f(x,y)$ is integrable, and $\partial{f}/\partial{x}(x,y)$ is bounded function of $(x,y)$. Show that $\partial{f}/\partial{x}(x,y)$ is a measurable function of $y$ for each $x$ and $$\frac{\partial}{\partial{x}}\int^{1}_{0} f(x,y)dy = \int^{1}_{0} \frac{\partial{f}}{\partial{x}}(x,y)dy$$ Assume that $\partial{f}/\partial{x}(x,y)$ exists everywhere for now (I'm not sure if we can assume this):

I think the problem is asking us to justify using Leibniz rule by applying dominated convergence theorem to the sequence of functions $$\phi_{h}(x,y) := \frac{f(x+h,y)-f(x,y)}{h}.$$ To do this, we need to show that the sequence is dominated by an integrable function, by the mean value theorem, there is an $x_{0} \in (x, x+h)$ where

$\phi_{h}(x,y)= \frac{\partial{f}}{\partial{x}}(x_{0},y)$

But I am a bit stuck relating $\frac{\partial{f}}{\partial{x}}(x_{0},y)$ to the integrable function $f$. I think I am missing some estimate.

You're overthinking this. Since $\partial{f}/\partial{x}(x,y)$ is bounded on all of $[0,1]\times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
$\left\lvert\frac{\partial f}{\partial x}(\xi_h,y)\right\rvert\le \sup\limits_{\text{domain}}\left\lvert\frac{\partial f}{\partial x}\right\rvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$\lim_{h\to 0}\int_0^1\phi_h(x,y)\,dy=\int_0^1\lim_{h\to 0}\phi_h(x,y)\,dy$$