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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
up vote 1 down vote accepted

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?

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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
I'm now trying to think about the sequence of functions assumed in the Lebesque DCT's assumptions. I suppose the Newton quotient would be the converging sequence? – KUSH Dec 4 '12 at 6:25
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
Indeed. Thanks a bunch! – KUSH Dec 4 '12 at 7:11
@MhenniBenghorbal I have a question on how to use LDCT for this example. We see that $\lim_{n\rightarrow \infty} \dfrac {f(x,y+1/n)-f(x,y)}{1/n} = \dfrac {\partial f} {\partial y}$. But we are only given that $\left| \dfrac {\partial f} {\partial y}(x,y) \right| \le g(x)$ so how do we know $ \left| \dfrac {f(x,y+1/n)-f(x,y)}{1/n} \right| $ is dominated by an integrable function – n.e. Jun 6 at 0:00

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