$a_n=\frac{1}{n}\int_{0}^{\infty}e^{-\frac{t}{n}}f(t)dt\in\mathbb{R}$, then $a_n\rightarrow 1$ as $n\rightarrow\infty$

Question

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be Lebesgue inegrable and assume that $f(t)\rightarrow 1$ as $t\rightarrow\infty$. Prove that for each positive integer $n$ we may define $a_n=\frac{1}{n}\int_{0}^{\infty}e^{-\frac{t}{n}}f(t)dt\in\mathbb{R}$ and prove that $a_n\rightarrow 1$ as $n\rightarrow\infty$.

Answer 1

Note that $$ \frac{1}{n}\int_0^\infty e^{-\frac{t}{n}}dt = -e^{-\frac{t}{n}}|_0^{\infty} = 1.$$ Since $f(t) \to 1$ as $t \to \infty$, choose $T$ such that for $t > T$, $$ |f(t) - 1| < \epsilon.$$ Then assuming $f$ is absolutely integrable, i.e., $\|f\|_{L_1} < \infty$, \begin{align} |a_n - 1| &\le \frac{1}{n}\left(\int_0^T e^{-\frac{t}{n}}|f(t)-1|dt + \int_T^\infty e^{-\frac{t}{n}}|f(t)-1|dt \right) \\ &\le \frac{1}{n}\left(T + \|f\|_{L_1} + ne^{-\frac{T}{n}}\epsilon \right) \to \epsilon \quad \text{as } n \to \infty. \end{align}

To be more precise, \begin{align} \int_0^T e^{-\frac{t}{n}}|f(t)-1|dt &\le \int_0^T |f(t)|dt + \int_0^T 1dt \\ &\le \int_0^\infty |f(t)|dt + T = \|f\|_{L_1} + T. \end{align}