Pointwise and uniform convergence of sequence of functions Let $(f_n)$ be a sequence of continuous functions on $\mathbb R$.


*

*If $(f_n)$ converges to $f$ pointwise on $\mathbb R$, then $$\lim\limits_{n\to \infty}\int^{\infty}_{-\infty}f_n(x)dx=\int^{\infty}_{-\infty}f(x)dx.$$

*If $(f_n)$ converges to $f$ uniformly on $\mathbb R$ then $$\lim\limits_{n\to \infty}\int^{\infty}_{-\infty}f_n(x)dx=\int^{\infty}_{-\infty}f(x)dx.$$

*If $(f_n)$ converges to $f$ uniformly on $\mathbb R$ then $f$ is continuous on $\mathbb R$.

*There exists a sequence of continuous functions $(f_n)$ on $\mathbb R$, such that $f_n$ converges to $f$ uniformly on $\mathbb R$, but $$\lim\limits_{n\to \infty}\int^{\infty}_{-\infty}f_n(x)dx\neq\int^{\infty}_{-\infty}f(x)dx.$$
The option 3. is definitely correct and option 1. is definitely wrong. But I am in confusion about 2 and 4. please help!
 A: Here's an example that will show 1 and 2 false, and 4 correct. The idea is to have a bump that gets wider but smaller, keeping the same surface area.
Define the function $g$ on $\mathbb{R}$ by
$$\forall x\in\mathbb{R},\ g(x)=\frac1{1+x^2}.$$
It is well-known that $g$ is continuous on $\mathbb{R}$, that
$$\int_{-\infty}^{+\infty}g(x)\,\mathrm{d}x=\pi$$
and
$$\lVert g\rVert_\infty=1.$$
Define the sequence of functions $(f_n)_{n\in\mathbb{N}^*}$ on $\mathbb{R}$ as
$$\forall n\in\mathbb{N}^*,\ \forall x\in\mathbb{R},\ f_n(x)=\frac1ng\left(\frac xn\right).$$
Then, clearly, for all $n\in\mathbb{N}^*$, $f_n$ is continuous on $\mathbb{R}$,
$$\lVert f_n\rVert_\infty=\frac1n\lVert g\rVert_\infty=\frac1n$$
and hence
$$\lim_{n\to+\infty}\lVert f_n\rVert_\infty=0$$
hence the sequence of functions $(f_n)_{n\in\mathbb{N}^*}$ converges uniformly (and hence pointwise) to the nil function $f=0$ on $\mathbb{R}$.
Now, for all $n\in\mathbb{N}^*$,
$$\int_{-\infty}^{+\infty}f_n(t)\,\mathrm{d}t=\int_{-\infty}^{+\infty}\frac1ng\left(\frac tn\right)\,\mathrm{d}t\overset{(*)}=\int_{-\infty}^{+\infty}g(x)\,\mathrm{d}x=\pi$$
where for $(*)$ we used the substitution $x=\dfrac tn$.
Hence:
$$\lim_{n\to+\infty}\int_{-\infty}^{+\infty}f_n(t)\,\mathrm{d}t=\pi\neq\int_{-\infty}^{+\infty}f(t)\,\mathrm{d}t=0.$$
