Derivative of Integral = Integral of Derivative

[e(ho0n3]

Homework Statement

Let f : [0,1] x [0,1] -> R be a bounded function such that for each fixed y, the function g given by g(x) = f(x,y) is measurable. Suppose \partial f / \partial y exists and is bounded. Prove that

\frac{d}{dy} \int_0^1 f(x,y) \, dx = \int_0^1 \frac{\partial f}{\partial y} \, dx

Homework Equations

This is from the chapter discussing the Lebesgue convergence theorem: Let g be an integrable function (integrable meaning the Lebesgue integral is finite) over E and let (f_n) be a sequence of measurable functions that converge almost everywhere to f and |f_n| < g. Then

\int_E f = \lim \int_E f_n

Corollary: Let (g_n) be a seq. of integrable functions which converge a.e. to an integrable function g. Let (f_n) be a seq. of measurable functions such that |f_n| \le g_n and f_n \to f a.e. If \int g = \lim \int g_n, then \int f = \lim \int f_n.

The Attempt at a Solution

I need a sequence (h_n) such that h_n \to \partial f / \partial y and such that \lim \int h_n = d/dt \int_0^1 f(x,y) \, dx. I have no idea about the former. For the latter, since the derivative is a limit, I figure that we can use

h_n(y) = \int_0^1 \frac{f(x, y+a_n) - f(x,y)}{a_n} \, dx

where (a_n) is any sequence that converges to 0. I can't think of anything else at the moment. Any tips?

 

[rochfor1]

You're on the right track...hit this with the dominated convergence theorem!

 

[e(ho0n3]

To use dominated convergence, I need to prove that (h_n) is point-wise bounded. Since f is a bounded function, f(x,y + a_n) - f(x,y) is bounded. The problem is the 1/a_n factor of h_n, which is unbounded.