I am struggling with a question in Royden's book 4-edition, in p.90 #36
Let $f$ be a real-valued function of two variables $(x, y)$ that is defined on the square $Q = \{(x, y) : 0 \leq x \leq 1, 0 \leq y \leq 1\}$ and is a measurable function of $x$ for each fixed value of $y$. For each $(x,y) \in Q$ let $\partial f/ \partial y$ exists, suppose further $$|\frac{\partial f} {\partial y} (x,y)|\leq g(x)$$ for all $(x,y)\in Q$, and $g$ is an integrable function on [0,1]. Prove that $$\frac{d}{dy}\int_0^1f(x,y)dx=\int_0^1 \frac{\partial f}{\partial y}(x,y) dx$$ for all $y \in [0,1]$.
Here is my attempt.
Let $y \in [0,1]$. For each positive integer $n$, define $f_n(x):=f(x,y+y_n)$ such that ${y_n}\subseteq [0,1]$ and $y_n \rightarrow 0$ as $n \rightarrow \infty.$ Then, $f_n(x) \rightarrow f(x,y)$ uniformally, so pointwise on $[0,1]$. By assumption $\partial f_n(x) /\partial y$ exists for all $n,$ but that does not gurantee that $\partial f_n(x) /\partial y \rightarrow \frac{\partial f}{\partial y}(x,y)$ pointwise, if I am right.
The idea is to use the Lebesgue Dominated Convergence Theorem which I am not able to apply here, so I would appreciate any help with that
Take a sequence $\{h_n\}$ that converges to $0$. Then by the definition of the derivative and the linearity of the integral, \begin{align} \frac{\mathrm d}{\mathrm dy} \int_0^1 f(x,y) \, dx &= \lim_{n\to\infty} \frac{\int_0^1 f(x,y+h_n) \, dx - \int_0^1 f(x,y) \, dx}{h_n}\\ &= \lim_{n \to \infty} \int_0^1 \frac{f(x,y+h_n)-f(x,y)}{h_n} \, dx. \end{align} The Mean Value Theorem combined with the assumption shows that the integrand is dominated by an integrable function, so that you may invoke the DCT to interchange the limit and the integral.