# Under given conditions whether $\lim\limits_{n\to \infty} \int_{-\infty}^{\infty}f_n(t)dt=\int_{-\infty}^{\infty}f(t)dt$ or not?

Let $$\{f_n\}_{n=1}^{\infty}$$ be a sequence of continuous real-valued functions defined on $$\mathbb R$$ which converges pointwise to a continuous real-valued function $$f$$. Which of the following statements are true?

(i) If $$0\leq f_n \leq f$$ for all $$n\in \mathbb N$$ then $$\displaystyle \lim_{n\to \infty} \int_{-\infty}^{\infty}f_n(t)dt=\int_{-\infty}^{\infty}f(t)dt.$$

(ii) If $$|f_n(t)|\leq |\sin t|$$ for all $$t\in \mathbb R$$ and for all $$n\in \mathbb N,$$ then $$\displaystyle \lim_{n\to \infty} \int_{-\infty}^{\infty}f_n(t)dt=\int_{-\infty}^{\infty}f(t)dt.$$

(i) If $$\int_{-\infty}^{\infty} f<\infty$$, then we can use DOMINATED CONVERGENCE theorem and can say that the statement is true. But if $$\int_{-\infty}^{\infty}f=\infty$$ then what can we say about the statement?

(ii) I was not able to do this one.

Note: At the answer-key it's given that (i) is true but (ii) is false.

(1) Suppose $$0 \le f_n \le f$$ and $$\int_{-\infty}^\infty f(t)\; dt = \infty$$. Given $$N > 0$$, there is $$M$$ such that $$\int_{-M}^M f(t)\; dt > N$$. By dominated convergence $$\int_{-M}^M f_n(t)\; dt \to \int_{-M}^M f(t)\; dt$$, so $$\int_{-\infty}^\infty f_n(t)\; dt \ge \int_{-M}^M f_n(t)\; dt > N$$ for sufficiently large $$n$$.
(2) Try $$f_n(t) = \sin(t)$$ for $$n \pi < t < (n+1)\pi$$, $$0$$ otherwise.
• $f$ is stated to be a continuous real-valued function. – Robert Israel Jan 10 at 16:25