Uniform Convergence; Continuity

Question

In the last step. I believe we did

$$\frac{d}{dx} \int_{a}^{x} g = \frac{d}{dx} \lim_{n \to \infty} \int_{a}^{x} f_n' = \lim_{n \to \infty}\frac{d}{dx} \int_{a}^{x} f_n' = \lim_{n \to \infty} f_n'(x)$$

How come we are allow to switch between the limit operator and the differentiation operator?

Answer 1

Since $g$ is continuous, $\int_a^x g(t)\,dt$ is differentiable and its derivative is $g(x)$. (Fundamental theorem of calculus.)

So the last step is just differentiating the equality $$ f(x)-f(a) = \int_a^x g(t)\,dt$$ and using the earlier computation and the assumption that $f_n'$ converges (even uniformly) to $g$.

Answer 2

By the fundamental theorem of calculus $$\frac{d}{dx} \int_{a}^{x} g=g(x)$$But, by the first part, also $$\frac{d}{dx} \int_{a}^{x} g=\frac d {dx}[f(x)-f(a)]=f'(x)$$This is all.