# Real valued function of two variables defined on a square with area one, Partial derivatives exist and bounded by an Lebesque 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!

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Theorems from this section: (1) Lebesque Dominated Convergence Theorem. (2) General Lebesque Dominated Convergence Theorem. (3) Integral Comparison Test (4) MCT (5) Fatou's Lemma. –  KUSH Dec 4 '12 at 5:43
I think the Integral Comparison Test might be useful. Assume $f$ measurable, assume there is a nonnegative function dominated by an integrable function $g$ i.e. $|f|\le g$ on $E$. Then $|\int_E f| \le \int_E |f|$ –  KUSH Dec 4 '12 at 5:45

Here is how you start, Let $F(y)= \int_0^1 f(x,y)dx$, then
$$F'(y) = \lim_{h\to 0}\frac{F(y+h)-F(y)}{h}=\lim_{h\to 0} \int_0^1 \frac{(f(x,y+h)-f(x,y))}{h} dx .$$
I didn't think to examine $\frac {d} {dy}$ as a Newton quotient. Given that $\frac {\partial f} {\partial y}$ is dominated by $g$, the Dominated Convergence theorem would be my initial guess (and since the results states the limit movement outright). –  KUSH Dec 4 '12 at 6:19
Ahh, I think that this should work, $\lim_{n\rightarrow \infty} \frac {f(y+1/n)-f(y)}{1/n} = f'(y)$ for almost every y. –  KUSH Dec 4 '12 at 7:05