# Let $\{f_n\}^\infty_{n=1}$ be a sequence of continuous real valued functions defined on $\mathbb{R}$

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

a. If $0\le f_n \le f$ for all $n\in \mathbb{N}$, then

$$\lim_{n\to \infty} \int_\infty^{-\infty} f_n(t)\,dt=\int_\infty^{-\infty } f(t) \, dt$$

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

c. If $|f_n(t)|\le e^t$ for all $t\in \mathbb{R}$ and for all $n\in \mathbb{N}$, then for all $a,b \in \mathbb{R}$ . $a<b$

$$\lim_{n\to \infty} \int_b^a f_n(t) \, dt=\int_b^a f(t) \, dt$$

since sequence of continuous real valued functions defined on $\mathbb{R}$ which converges pointwise to a continuous real valued function $f$ so 1 is true (but ia m not sure) can you hlep me with other options too..thank you

• It seems to me that you are meant to know some form of the dominated convergence theorem, and that the exercise is an excuse for having you come up with counterexamples to some possible weaker formulations.
– user228113
Dec 31, 2017 at 13:43

For case b., consider the sequence $(f_n)$ where $f_n(t) = \vert \sin t \vert$ for $t \in [n \pi, (n+1)\pi]$ and vanishes elsewhere. $(f_n)$ converges pointwise to the always vanishing function.

However, you have for all $n \in \mathbb N$

$$2 = \int_{-\infty}^\infty f_n(t) \ dt \neq \int_{-\infty}^\infty 0 \ dt =0.$$

Cases a. and c. are consequences of Lebesgue dominated convergence theorem providing $\int_{-\infty}^\infty f(t) \ dt$ converges for case a. The theorem doesn’t apply to case b. as $\int_{-\infty}^\infty \vert \sin t \vert \ dt$ diverges.

• @mathcounterexamples.net...awesome i want your brain some times.. Dec 31, 2017 at 14:07
• @mathcounterexamples.net..is i am correct with a Dec 31, 2017 at 14:08
• @suresh I’d also like to be much stronger in math ;-). What you wrote about case a. is correct. It’s a direct use of Lebesgue dominated convergence theorem providing $\int_{-\infty}^\infty f(t) \ dt$ converges. It is also true if $\int_{-\infty}^\infty f(t) \ dt$ diverges providing the hypothesis $0 \le f_n \le f$. For the proof, you can consider the intervals $[-M,M]$ with $M >0$ converging to $\infty$. Dec 31, 2017 at 14:21

(a) and (c) are true. (b) is wrong.

- Proof (a). By Fatou's Lemma one has: \begin{align} \int_\mathbb{R} f(x)\,dx \leq \liminf_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx \end{align} Moreover we have by monotonicity of the integral: \begin{align} \limsup_{n\to\infty}\int_\mathbb{R} f_n(x)\,dx \leq \int_\mathbb{R} f(x)\,dx \end{align} Hence: \begin{align} \limsup_{n\to\infty}\int_\mathbb{R} f_n(x)\,dx \leq \int_\mathbb{R} f(x)\,dx \leq\liminf_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx \end{align} So: \begin{align} \lim_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx =\int_\mathbb{R}f(x)\,dx \end{align}

- Counterexample (b). Do it yourself. See for example @mathcounterexamples.net

- Proof (c). Since $e^t$ is integrable on $[a,b]$ the result follows from Dominated Convergence Theorem.

• Why downvote? I'm ready to correct mistakes. Dec 31, 2017 at 13:58
• Corrected the answer. Now okay? Dec 31, 2017 at 14:19