Let ${f_n}$ be a sequence of continuous real valued functions on $[0, \infty)$, then which of the following is/are true?

Question

Let $\{f_n\}$ be a sequence of continuous real valued functions on $[0, \infty)$. Suppose $f_n(x)\to f(x) ~~~\forall x\in [0,\infty)$ and $f$ is integrable. Then

1. $\int_{0}^{\infty} f_n(x)dx \to \int_{0}^{\infty} f(x)dx$ as $n\to \infty$.
2. if $f_n\to f$ uniformly on $[0,\infty)$ then $\int_{0}^{1} f_n(x)dx\to\int_{0}^{1} f(x)dx$.

3. if $f_n\to f$ uniformly on $[0,\infty)$ then $\int_{0}^{\infty} f_n(x)dx\to\int_{0}^{\infty} f(x)dx$.

4. if $\int_{0}^{1}|f_n(x)-f(x)|dx\to 0$ then $f_n\to f$ uniformly on $[0,1]$

That is the question. Now I have to admit that I don't have the slightest idea how to solve this. I know about convergence of sequences but don't know how to figure out when integrals involved. Any hint, reference or solutions will be nice. Or some books where I can learn them. Thanks for your help.

Answer 1

$(1)$ is false by the conterexample $f_n(x) = \frac1n\chi_{[1,n]}(x)$ which converges pointwise to $0$, but $\int_0^\infty f_n(x)\ \mathsf dx = 1$ while $\int_0^\infty 0\ \mathsf dx = 0$.

$(2)$ is true by a standard theorem in introductory real analysis - If a sequence of continuous functions converges uniformly on a closed interval, then the limit and integral can be interchanged. This follows from the $f_n$ being uniformly continuous and $f$ itself being continuous.

$(3)$ is false by the same counterexample as in $(1)$ (since $\sup_{x\in[0,\infty)}|f_n(x)|=\frac1n\stackrel{n\to\infty}\longrightarrow 0$).

$(4)$ is false as by the counterexample of the "typewriter sequence", demonstrated on Terry Tao's blog in example 4: https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/