Real valued function of two variables defined on a square with area one, Partial derivatives exist and bounded by an Lebesgue intergrable function This problem in my real analysis textbook has been, let's just say, troubling me. Here is the problem:

Let $f$ be a real-valued function of two variables $(x,y)$ that is defined on the square $D=\{(x,y): 0\le x \le 1, 0 \le y \le 1\}$ and $f$ is a measurable function of $x$ for each fixed value of $y$. For each $(x,y) \in D$ let the partial derivative $\frac {\partial f} {\partial y}$ exist. Also Suppose there is a function $g$ that is integrable over $[0,1]$ such that $\frac {\partial f} {\partial y}(x,y) \le g(x)$ for all $(x,y) \in D$. Prove that:
  $$ \frac d {dy} \bigg[ \int_0^1 f(x,y)dx \bigg]= \int_0^1 \frac {\partial f} {\partial y}(x,y)dx \ \ \ \ \forall y \in [0,1] $$

What I'm really having trouble with is wrapping my head around what is going on. I will list what I know:


*

*Each $f$ is measurable on all of the $x$ values when $ y$ is fixed.

*The partial derivative of $f$ w.r.t. $y$ exists for each point i.e. $\ \ \exists \frac {\partial f} {\partial y}(x,y)$

*The partial derivatives are bounded by an integrable function $g$ of $x$, integrable implies $\int_{[0,1]} g < \infty$


So essentially what I'm trying to show is that the derivative of the area w.r.t. $y$ can be rewritten as the area of a partial derivative of $f$ w.r.t. $y$. Honestly, I just have no idea where to begin. Some tips, hints, or proofs would be greatly appreciated!
 A: Like the comments and the answer suggest one way to show this is by Dominated Convergence Theorem. Also we will make use of the following lemma from undergraduate level real analysis:
Lemma: Let $F:\mathbb{R}\to\mathbb{R}$ be a function and $x_0\in\mathbb{R}$. Then $$\lim_{x\to x_0}f(x)=L \iff [\forall \{x_n\}_n \subseteq \mathbb{R}: x_n\to x_0\implies f(x_n)\to L]. (\ast)$$
This lemma states that we can think of a continuous limiting process as a discrete limiting process, provided that we take into account any discretization of it. The utility of the lemma stems from the observation that if we have a discrete limiting process then we can define an appropriate sequence of functions.

Result: Let $f:[0,1]\times[0,1]\to\mathbb{R}$ be such that $\forall  y\in[0,1]:f(\cdot,y):[0,1]\to\mathbb{R}$ is measurable. Suppose $\forall (x,y)\in[0,1]\times[0,1]:\dfrac{\partial f}{\partial y}(x,y)$ exists, and that there is an integrable $g:[0,1]\to\mathbb{R}:$
$$\forall(x,y)\in [0,1]\times[0,1]: \left|\dfrac{\partial f}{\partial y}(x,y)\right|\leq g(x).$$
Then
$$\forall y\in[0,1]:\dfrac{d}{dy}\left[\int_{[0,1]} f(x,y)dx\right] =\int_{[0,1]} \dfrac{\partial f}{\partial y}(x,y) dx.$$
Proof of the Result: Fix $y\in[0,1]$. Let $\{h_n\}_n\subseteq\mathbb{R}-0: h_n\downarrow0$ and set 
$$\forall n: f_n:[0,1]\to\mathbb{R}, f_n(x):=\dfrac{f(x,y+h_n)-f(x,y)}{h_n}.$$
Note that $\forall n: f_n$ is measurable since it is a linear combination of measurable functions. Since the partial derivative with respect to $y$ at any $(x,y)$ exists we have
$$\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} \dfrac{f(x,y+h_n)-f(x,y)}{h_n} \stackrel{(\ast)}{=} \lim_{h\to 0} \dfrac{f(x,y+h)-f(x,y)}{h} \stackrel{\tiny\mbox{def}}{=} \dfrac{\partial f}{\partial y} (x,y).$$
Thus $f_n(\cdot)\to \dfrac{\partial f}{\partial y}(\cdot,y)$ pointwise on $[0,1]$.
Let $x\in[0,1]$. Since $f_n(x)\to\dfrac{\partial f}{\partial y}(x,y), \exists N,\forall n\geq N: \left|f_n(x)-\dfrac{\partial f}{\partial y}(x,y)\right|<1$. Hence $\forall n\geq N:$
\begin{align}
|f_n(x)|-\left|\dfrac{\partial f}{\partial y}(x,y)\right|\leq \left|f_n(x)-\dfrac{\partial f}{\partial y}(x,y)\right|<1 \\
|f_n(x)|< \left|\dfrac{\partial f}{\partial y}(x,y)\right|+1 < g(x)+1 \in L^1([0,1]).
\end{align}
Thus we may assume, by discarding the first $N-1$ terms of $\{f_n\}_n$ if necessary (i.e. by considering the sequence $\{f_n\}_{n\geq N}$)  that $\{f_n\}_n$ is dominated by an integrable function. Then by Lebesgue's Dominated Convergence Theorem, $\dfrac{\partial f}{\partial y}(\cdot,y)$ is integrable on $[0,1]$ and
$$\int_{[0,1]}f_n(x)dx\to\int_{[0,1]}\dfrac{\partial f}{\partial y}(x,y)dx.$$
Since the sequence $\{h_n\}_n$ converging to $0$ was arbitrary we may apply the lemma above:
\begin{align}
\lim_{n\to\infty}\int_{[0,1]}f_n(x)dx 
&= \lim_{n\to\infty}\int_{[0,1]}\dfrac{f(x,y+h_n)-f(x,y)}{h_n} dx \\
&=\lim_{n\to\infty}\dfrac{\int_{[0,1]}f(x,y+h_n)dx- \int_{[0,1]} f(x,y)dx}{h_n}\\
&\stackrel{(\ast)}{=} \lim_{h\to0}\dfrac{\int_{[0,1]}f(x,y+h)dx- \int_{[0,1]} f(x,y)dx}{h}
=\dfrac{d}{dy} \int_{[0,1]}f(x,y)dx.
\end{align}
Finally since $y$ is arbitrary the result holds for any $y$.

P.S.: For reference purposes this is Exercise 4.4.36 of Royden & Fitzpatrick's Real Analysis, 4e (pp. 90-91).
A: Here is how you start, Let $F(y)= \int_0^1 f(x,y)dx $, then
$$ F'(y) = \lim_{h\to 0}\frac{F(y+h)-F(y)}{h}=\lim_{h\to 0} \int_0^1 \frac{(f(x,y+h)-f(x,y))}{h} dx . $$
Now, you can see that, the problem is nothing, but interchanging limit with integral. So, what theorem you think you need? 
