The following question is from Real Analysis by Royden (4th Edition). More precisely, it is problem 36 of chapter 4. Further, the problem follows right after the section over The Lebesgue Dominated Convergence Theorem, hence, that is what should be used. Below is the problem followed by my attempted solution which I have a question about.
Problem:
Let $f$ be a real-valued function of two variables $\left(x,y\right)$ that is defined on the following unit square
$Q:=\{\left(x,y\right): 0\leq x\leq 1, 0\leq y\leq 1\}$ and is a measurable function of $x$ for each fixed value of
$y$. For each $\left(x,y\right)\in Q$ let the partial derivative a $\partial f/\partial y$ exist. Suppose there is a function $g$ that is integrable over $[0,1]$ and such that,
$$\left\vert\frac{\partial f}{\partial y}\left(x,y\right)\right\vert \leq g(x) \text{ for all } \left(x,y\right)\in Q.\tag{*}$$
Prove that
$$\frac{d}{dy}\left[\int_0^1 f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx, \text{ for all } y\in[0,1].$$
(My) Proof:
Fix $y\in[0,1]$ and consider the sequence of functions $f_n\colon[0,1]\to\mathbb{R}$ defined by,
$$f_n(x):=\frac{f(x,y+1/n) – f(x,y)}{1/n}.$$
Then, $\{f_n\}$ is a sequence of measurable functions ($f$ is a measurable function of $x$ for each fixed value of $y$) dominated by $g(x)$ that converge pointwise almost everywhere to $\partial f/\partial y$ on $[0,1]$ (for each $(x,y)\in Q$ the partial derivative $\partial f/\partial y$ exists). Therefore, by The Lebesgue Dominated Convergence Theorem,
$$\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$$
$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) – f(x,y)}{1/n}dx =
\int_0^1\frac{\partial f}{\partial y}(x,y)dx.\tag{1}$$
Notice,
$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) – f(x,y)}{1/n}dx =
\lim_{h\to 0}\frac{\int_0^1f(x,y+h)dx-\int_0^1f(x,y)dx}{h}=
\frac{d}{dy}\left[\int_0^1f(x,y)dx\right].\tag{2}$$
Thus, combining (1) and (2),
$$\frac{d}{dy}\left[\int_0^1f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx,$$
which was to be shown.
My main concern with my proof is that, I am not sure how the sequence $\{f_n\}$ are dominated by $g(x)$. I know by (*) that the limit of the sequence $\{f_n\}$ is bounded by $g(x)$. Also, if there are any mistakes in the proof, please let me know!
Edit:
Does the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence?
Best Answer
Consider $y \in [0,1]$ and any sequence $y_n \in [0,1]$ such that $y_n \to y$. By the mean value theorem, there exists $\xi_{x,n}$ between $y$ and $y_n$ such that
$$|f_n(x)| = \left|\frac{f(x,y_n) - f(x,y)}{y_n - y} \right| = \left|\frac{\partial f}{\partial y}(x,\xi_{x,n})\right| \leqslant g(x)$$
We have
$$\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y} = \frac{\partial f}{\partial y}(x,y),$$
and by the dominated convergence theorem
$$\lim_{n \to \infty}\int_0^1\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1 \frac{\partial f}{\partial y}(x,y) \, dx$$