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Consider all sequences $\{f_n\}$ of real valued continuous functions on $[0,\infty)$. Identify the correct options.

  1. If $\{f_n\}$ converges to $f$ pointwise on $[0,\infty)$, then $$\lim_{n\to \infty} \int^{\infty}_0 f_n(x) dx\,=\int^{\infty}_0 f(x) dx\,$$

  2. If $\{f_n\}$ converges uniformly to $f$ on $[0,\infty)$, then $$\lim_{n\to \infty} \int^{\infty}_0 f_n(x) dx\,=\int^{\infty}_0 f(x) dx\,$$

  3. If $\{f_n\}$ converges to $f$ uniformly, then $f$ is continuous on $[0,\infty)$.

  4. There exists a sequence of continuous functions $\{f_n\}$ on $[0,\infty)$ such that $\{f_n\}$ converges to $f$ uniformly on $[0,\infty)$ but $$\lim_{n\to \infty} \int^{\infty}_0 f_n(x) dx\,\neq \int^{\infty}_0 f(x) dx\,$$

I think options 3,4 are correct, for 1) a simple counterexample can be constructed of triangles by joining the points (0,0), (0,1) and (1/n,0). Proving 3 is also same as for the case of bounded intervals. But I don't know how to give counter for 2. and how to provide example for 4.

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    $\begingroup$ Should the right hand sides be $\displaystyle\int_{0}^{\infty}f(x)\,dx$ (without the $n$)? $\endgroup$
    – JimmyK4542
    Dec 28 '15 at 9:46
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For 1, you sorta have the right idea, but you need to be careful with how you construct your counterexample. By "joining the points (0,0), (0,1) and (1/n,0)." you get a triangle with base $\tfrac{1}{n}$ and height $1$, which has an area of $\tfrac{1}{2n}$. Thus, $\displaystyle\int_{0}^{\infty}f_n(x)\,dx = \dfrac{1}{2n} \to 0 = \int_{0}^{\infty}f(x)\,dx$ as $n \to \infty$. Instead, try joining the points $(0,0)$, $(\tfrac{1}{n},n)$, and $(\tfrac{2}{n},0)$. Then you will get a triangle with base $\tfrac{2}{n}$ and height $n$, which has an area of $1$. Hence, $\displaystyle\int_{0}^{\infty}f_n(x)\,dx = 1$ for all $n$ while $\displaystyle\int_{0}^{\infty}f(x)\,dx = 0$.

For 2 and 4, consider something like $f_n(x) = \dfrac{1}{n}$ for all $x \in [0,\infty)$.

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