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!
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?