Let $f:[0,1]^2\to\mathbb{R}$ be a bounded funtion. Suppose that for each fixed $y$, the map $x\to f(x,y)$ is a measurable function of $x$. Suppose the partial derivative $\frac{df}{dy}$ exists at every point of $[0,1]^2$ and defines a bounded function o $[0,1]^2$. Prove that:
$$\frac{d}{dy}\int\limits_{0}^1f(x,y)dm(x)=\int\limits_0^1\frac{df}{dy}dm(x)$$.
My attempt:
$\begin{align}
\frac{d}{dy}\int\limits_{0}^1f(x,y)dm(x) &=\lim\limits_{h\to0}\frac{\int\limits_0^1f(x,y+h)-\int\limits_0^1f(x,y)}{h}\\
&=\lim\limits_{h\to0}\int\limits_0^1\frac{f(x,y+h)-f(x,y)}{h}
\end{align}$
We have done several results to interchange the limit and integral when you have a sequence of functions converging pointwise almost everywhere. So to take the limit inside the integral, I'm going to use the fact that:
$\lim\limits_{x\to a}f(x)=b$ when ever for any sequence $x_n$ with $\lim\limits_{n\to\infty}x_n=a$, then $\lim\limits_{n\to\infty}f(x_n)=b$
This is the place I got stuck:
So may be I can define $g(h)=\frac{f(x,y+h)-f(x,y)}{h}$ and for $(h_n)$ such that $h_n\to0$ $g_n=g(h_n)$ ?or I might be wrong.
Appreciate your help to complete the proof
\begin{align*} \left|\dfrac{f(x,y+h)-f(x,y)}{h}\right|&=\left|\dfrac{\partial f}{\partial y}(\eta_{x,y,h})\right|\leq M \end{align*} as the assumption is given that the partial is bounded, here $\eta_{x,y,h}$ is chosen by Mean Value, and we are now to use Lebesgue Dominated Convergence Theorem.