I am working on Royden's Real Analysis 4th edition, Chapter 4.4, #36.
Please let me know how I should go forth:
Let $f(x,y)$ be a real-valued function defined on a square $Q=\{(x, y) | 0 \leq x, y \leq 1\}$ and is a measurable function of $x$ for each fixed value of $y$. For each point $(x, y) \in Q$ let $\frac{\partial f}{\partial y}$ exist. If there is a function $g$ integrable over $\lbrack 0, 1 \rbrack$ such that $\vert \frac{\partial f}{\partial y} (x, y) \vert \leq g(x)$ for all $(x, y) \in Q$, show that $\frac{d}{dt}\lbrack \int_0^{1} f(x, y) \,dx \rbrack = \int_0^{1} \frac{\partial f}{\partial y}(x, y) dx$ for all $y \in \lbrack 0, 1\rbrack$.
Thank you very much!
Here is a sketch of a proof, leaving you a few details to fill in.
The function $x \to f(x,y)$ is integrable for each $y$.
Consider a sequence $(y_n)$ such that $y_n \to y.$
By the Mean Value Theorem, there is some $\xi$ between $y_n$ and $y$ such that $$\left| \frac{f(x,y_n) -f(x,y)}{y_n-y} \right| = \left|\frac{\partial f}{\partial y}(x,\xi) \right| \leqslant g(x).$$
Now apply the Dominated Convergence Theorem to
$$\frac{F(y_n)-F(y)}{y_n -y}= \int_0^1 \frac{f(x,y_n)-f(x,y)}{y_n-y} \, dx,$$
where
$$F(y) = \int_0^1 f(x,y) \,dx.$$