Uniform convergence of integral on unbounded interval

Question

Suppose I have $\;f_n:[a,\infty)\to\mathbb{R}$ and $\int_a^\infty f_n(x) dx$ exists. If $f_n\to f$ uniformly on $[a,\infty)$, I am able to show that $\int_a^\infty f(x)dx$ exists and $\int_a^\infty f(x)dx = \lim\limits_{b\to\infty}\lim\limits_{n\to\infty}\int_a^b f_n(x)dx$

Now the question is $$g_n(b):= \int_a^b f_n(x)dx,\;\;g(b) = \int_a^b f(x)dx$$

I know that $g_n(b)\to g(b)$ uniformly on any interval $[a,M]$, but is it true that

$$\lim_{n\to\infty}\lim_{b\to\infty}g_n(b) = \lim_{b\to\infty}\lim_{n\to\infty}g_n(b)$$

This is my attempt in trying to prove (disprove) $\int_a^\infty f(x)dx =\lim\limits_{n\to\infty}\int_a^\infty f_n(x)dx$, please help me.

Answer 1

$$f_n(x) = \left\{\begin{array}{l l} \frac{1}{x} & \quad x\le n\\ \frac{1}{x^2} & \quad x>n\end{array}\right.\hspace{3cm}f(x) = \frac{1}{x}$$
Proof: $f_n(x)\to f$ on $[1,\infty)$
$$\forall \epsilon>0.\;\exists \frac{1}{n}<\epsilon.\;\sup\limits_{x\in [1,\infty)}|f_n(x)-f(x)| =\sup\limits_{x\in [n,\infty)}|\frac{1}{x}-\frac{1}{x^2}|<\sup\limits_{x\in [n,\infty)}|\frac{1}{x}|<\epsilon$$
$\Longrightarrow f_n(x)\to f(x)$ uniformly on $[1,\infty)\hspace{1cm}//$

$$b>n.\;\int_1^b f_n(x) dx=\log(n)+\frac{1}{n}-\frac{1}{b}\Longrightarrow \int_1^\infty f_n(x)dx = \log(n)+\frac{1}{n}$$

$\int_1^b f(x)dx = \log(b)\Longrightarrow\int_1^\infty f(x)dx$ does not exist.

So my statement is false in my question. Sorry about that.

Answer 2

Let $f (x) = \frac 1 x$ and $f_n(x) = \frac 1 x 1 \{ x < n \}$. Then all assumptions are satisfied but we don't have $\int_a^\infty f(x) dx < \infty$.