Let $f: [0, \infty) \to \mathbb R$ be Lebesgue integrable and assume that $f(t)\to 1$ as $t\to \infty$. Prove that for each positive integer $n$ we may define \begin{equation} a_n = \frac{1}{n} \int_0^\infty e^{-t/n} f(t)dt \in \mathbb R \end{equation} and prove that $a_n \to 1$ as $n \to \infty$.
my attempt
Since $0 < \frac{1}{n}e^{-t/n} \leq 1$ on $[0, \infty)$, the existence of each $a_n$ follows from the fact that $f$ is Lebesgue integrable (and hence absolutely integrable). \begin{align} &|a_n - 1| \leq \frac{1}{n} \int_0^\infty e^{-t/n} |f(t) - 1| dt \leq \frac{1}{n}\epsilon \int_a^\infty e^{-t/n} dt + \frac{1}{n} \int_0^a e^{-t/n}|f(t) - 1| dt, \end{align} where $\epsilon > 0$ and $a$ are chosen such that $|f(t) - 1| \leq \epsilon$ whenever $t \geq a$, which is possible by convergence of $f(t)$ to $1$. \begin{align} |a_n - 1| \leq \frac{1}{n}\epsilon \int_a^\infty e^{-t/n} dt + \frac{1}{n} \int_0^a|f(t) - 1| dt \leq \frac{1}{n}\epsilon \int_a^\infty e^{-t/n} dt + \frac{1}{n}M, \end{align} where $M$ is some constant.
Then the second term goes to $0$ as $n$ goes to $\infty$.
question 1
But what is a good way to bound the first term such that it converges to $0$ with $n\to \infty$?
answer 1
As Daniel Fischer pointed out in the comments below, one can substitute $u = \frac{t}{n}$ and the first term would integrate to $\epsilon e^{-a/n}$.
Then we have \begin{equation} \limsup_{n\to \infty} |a_n - 1| \leq \lim_{n\to \infty} \bigg(\epsilon e^{-a/n} + \frac{1}{n}M\bigg) = \epsilon. \end{equation} And since the choice of $\epsilon$ is arbitrary, $0 \leq \limsup_{n \to \infty}|a_n - 1| \leq 0$, which implies $a_n \to 1$.
question 2
I am curious how $f$ can be Lebesgue integrable on $[0, \infty)$ when it converges to $1$. The measue of the set is infinite but the function does not tend to $0$ and the function is still integrable, could you please give an example? Many thanks.
answer 2
As Calvin Khor and Daniel Fischer pointed out in the comment, such an $f$ cannot be Lebesgue integrable. But a way to make sense of the question is that $f$ is integrable on $[0, z]$ for every $z \in (0, \infty)$.
Many thanks to Daniel Fischer and Calvin Khor for their very helpful comments!
ad question 1: First note that there is no loss of generality to assume that $\epsilon<1$ and $a>1$ (why?). Then simply compute the integral in the first term. You should get (unless I made a mistake, which is not unlikely) $\epsilon e^{-a/n}\leq e^{-n}$ (why?), which then goes to zero.
ad question 2: Let's assume for starters that $f$ is measurable. Clearly $f$ could not be Lebesgue integrable on the full real line as it is bounded below by $1-\epsilon>0$ in $(a,\infty)$. It is, however, locally integrable on some interval $[A,\infty)$ (why?). Fortunately, you do not need need the full integrability assumption, you can just assume that $f$ is locally integrable on $[0,\infty)$ (so that you can compute $\int_0^a|f-1|=M<\infty$) and tends to $1$ at $\infty$ and the problem is well posed, as per your solution and my answer for question 1.