A uniformly convergent sequence of real valued functions need not to be term by term Integrable? We know that if $\{f_n\}_{n=1}^\infty$ be a  sequence of real valued functions converging uniformly to a function $f$ in a bounded interval $[a, b]$, then sequence is term by term integrable, i. e.,
$$\lim_{n\to\infty}\int_a^bf_n(x)dx= \int_a^b\lim_{n\to\infty}f_n(x)dx= \int_a^bf(x)dx$$
I am looking for the case when interval is unbound. I knew that result is not true for unbounded interval. But, I am anable to find/construct a counter example. Suggest me how can I find an example to disprove this result in unbounded intervals.
 A: What you need to know ...vs... What you  want to know.
A simple counterexample shoots down a possible conjecture but does not do much to instruct you about the topic itself, especially one as important as this one.
So ...
Situation for the Riemann integral with uniform convergence:

*

*[Uniform convergence]  Let $\{f_n\}$ be a sequence of Riemann integrable functions on an interval $[a,b]$  such that  $f_n\to f$ uniformly on $[a,b]$.  Then (as you say)
(i)  $f$ is also Riemann integrable on $[a,b]$, and (ii) the limit identity holds:
$$\lim_{n\to \infty} \int_a^bf_n(x)\,dx = \int_a^b \lim_{n\to \infty}f_n(x)\,dx =\int_a^bf(x)\,dx .\tag{1}$$
Situation for the Riemann integral with  pointwise convergence:


*[Pointwise convergence]  Let $\{f_n\}$ be a sequence of Riemann integrable functions on an interval $[a,b]$  such that  $f_n\to f$ pointwise [i.e., not uniformly] on $[a,b]$.  Then  (i)  $f$ might not be bounded and so it is not Riemann integrable on $[a,b]$, and (ii) even if $f$ is bounded it still might not be Riemann integrable so you cannot claim the limit (1).


*[Arzelà-Osgood Bounded Convergence Theorem]   Let $\{f_n\}$ be a sequence of Riemann integrable functions on an interval $[a,b]$  such that  $f_n\to f$ pointwise [i.e., not uniformly] on $[a,b]$. Suppose that  $\{f_n\}$ is uniformly bounded (i.e., there is a number $M$ so that $|f_n(x)|\leq M$ for all $n$ and all $x$.
Then
(i)  $f$ must  be bounded but it might not be Riemann integrable on $[a,b]$, and
(ii)  if, however,  $f$ is   Riemann integrable, then
$$ \int_a^bf_n(x)\,dx \to \int_a^b f(x)\,dx.$$
The unfortunate situation for the improper Riemann integral even with uniform convergence:
Let $\{f_n\}$ be a sequence of functions on an interval $[0,\infty)$  such that  $f_n\to f$ uniformly on   $[0,\infty)$.  Suppose that each integral $\int_0^\infty f_n(x)\,dx$ exists in the improper Riemann sense.
Then we know for sure (from the first part) that
(i)  The function $f$ must be Riemann integrable on each bounded interval $[0,T]\subset [0,\infty)$.
(ii)  For each bounded interval $[0,T]\subset [0,\infty)$
$$\lim_{n\to \infty} \int_0^T f_n(x)\,dx = \int_0^T \lim_{n\to \infty}f_n(x)\,dx = \int_0^T  f(x)\,dx .\tag{2}$$
But we do not know if
$$ \lim_{n\to \infty} \int_0^\infty f_n(x)\,dx
=  \int_0^\infty f(x)\,dx .\tag{3}$$
because that is the same as swapping a double limit:
$$ \lim_{n\to \infty} \int_0^\infty f_n(x)\,dx
= \lim_{n\to \infty} \lim_{T\to \infty}\int_0^T f_n(x)\,dx= \lim_{T\to \infty} \int_0^T \lim_{n\to \infty}f_n(x)\,dx = \int_0^\infty f(x)\,dx .$$
WARNING: Never, ever swap two limits $\lim_A\lim_B = \lim_B\lim_A$ without serious thinking. In this case there are three limits, since the integration itself is defined as  limit.
So the first step in understanding this is to search for counterexamples:

Problem.  In each of the three cases below find an example of a  uniformly convergent sequence   $\{f_n\}$  of functions on an interval
$[0,\infty)$  such that  $f_n\to f$ uniformly on   $[0,\infty)$ and
each integral $\int_0^\infty f_n(x)\,dx$ exists in the improper
Riemann sense but where
Part (a): The integral $\int_0^\infty f(x)\,dx $ does not exist in the
improper Riemann sense.
Part  (b):  The limit $\lim_{n\to \infty} \int_0^\infty f_n(x)\,dx$
does not exist.
Part (c):  The integral $\int_0^\infty f(x)\,dx $ does  exist in the
improper Riemann sense and the limit $\lim_{n\to \infty} \int_0^\infty
  f_n(x)\,dx$  also exists but the two are not equal.

In the comments is the example $f_n(x)= \frac1n$ for $0\leq x \leq n$ and $f_n(x)=0$ for $n<x\leq \infty$.  This sequence converges uniformly to the zero function but
$\int_0^\infty f_n(x)\,dx = 1$ does not converge to zero.  This answers question (c) and you have left for your amusement questions (a) and (b).
A: For $n=1,2, \ldots $, define $f_n:[0,\infty) \to \mathbb{R}$ by
$$f_n(x):= \frac{1}{n}\displaystyle{\mathbf 1}_{[0,n]}(x)+\frac{1}{2}\left(  \frac{3}{n}-\frac{x}{n^2}\right)\displaystyle{\mathbf1}_{(n,3n)}(x) \,,$$
where $\mathbf{1}_S$ (for $S\subseteq \mathbb{R }$) is the characteristic function
$$\mathbf 1_S (x) = \begin{cases}
1 & \text{ if } x \in S \\
0 &  \text{ if } x \notin S
\end{cases}$$
Consider the sequence  $\{f_n\}_{n=1}^\infty$. Notice $f_n \to 0$ uniformly as we have $$\sup_{x \in [0,\infty)} |f_n(x)|=\frac{1}{n} \quad (n=1,2,\ldots).$$
However notice that for $n=1,2,\ldots$, we have
$$\int_0^\infty f_n(x)\,dx = 2. $$
We may combine our observations to see that we apparently have
$$\lim_{n\to \infty} \int_0^\infty f_n(x)\,dx
= 2 \neq 0= \int_0^\infty \lim_{n\to \infty} f_n(x)\,dx .
$$

By linearity, the sequence given by the equation $\{2g_n\}_{n=1}^\infty:=\{f_n\}_{n=1}^\infty$ would be an example of a sequence that converges uniformly to $0$ but the limit of the integrals equals $1$.
