modern analysis: limits, integrals, uniform

Question

Suppose $\{f_n\} \to f$ uniformly on $[a, b]$ and both $f$ and the $f_n$ are integrable. Prove that $\lim_{n\to \infty}\int_{a}^bf_n(x)dx = \int_{a}^bf(x)dx$

Answer 1

Let epsilon e > 0 be given, choose N such that: / f(n)(x) - f(x) / < e/(b - a) for n > N and all x in [a, b]. Then:

/ Int(a to b)f(n)(x)dx - Int(a to b)f(x)dx / = / Int(a to b)( f(n)(x) - f(x))dx/ < (b - a)*e/(b - a) = e, and the result follows.

Answer 2

Hint

Use the uniform convergence of $\{f_n\}$ to $f$ and the inequality $$\left|\int_a^b f_n(x)dx-\int_a^b f(x)dx\right|\le\int_a^b||f_n-f||_\infty dx=(b-a)||f_n-f||_\infty$$

Answer 3

In all these limit integral interchanging the main idea of the proof goes into proving that it is integrable. For example once the function f to which series converges is riemann integrable then we can easily prove that the integral is limit of integral.

http://www.math.dartmouth.edu/~m43s12/notes/class8.pdf is one proof of the result.