Boundedness of sequence of functions with continuity and nonnegativity with convergent improper integral

Question

Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of continuous nonnegative functions in $[a,\infty)$ for some $a\in\mathbb{R}$. Moreover, we assume that $$\int_{a}^{\infty}f_{n}(t)dt\to 0 \text{ as }n\to\infty$$ Can we prove that $(f_{n})_{n\in\mathbb{N}}$ is uniformly bounded with respect to $t\in[a,\infty)$ and $n\in\mathbb{N}$? In other words, I want to show that $$\exists M>0 \,\forall n\in\mathbb{N} \,\forall t\in[a,\infty)\, |f_{n}(t)|\leq M$$

This is my attempt so far:
First, I use continuity of $f_{n}$ and assume the conclusion does not hold for the sake of contradiction. Then, I know that there exists a sequence $(f_{n}(t_{n}))_{n\in\mathbb{N}}$ such that $$\forall n\in\mathbb{N}\text{ (large enough) }, \exists \eta_{n}>0 \forall t\in(t_{n}-\eta_{n},t_{n}+\eta_{n}), f_{n}(t)\geq \frac{1}{2}$$ Therefore, for $n$ sufficiently large, we deduce that $$\eta_{n}\leq \int_{t_{n}-\eta_{n}}^{t_{n}+\eta_{n}}f(t)dt\leq \int_{a}^{\infty}f_{n}(t)dt\to0\text{ as }n\to\infty$$

The problem is that I cannot take $\eta_{n}$ uniformly to ensure $\inf\limits_{n\in\mathbb{N}}\eta_{n}>0$. Any help will much appreciated! Thank you!

Answer 1

Let $f_n(x)=n^{3}(x-n)$ for $n \leq x \leq n+\frac 1 {n^{2}}$, $f_n(x)=n^{3}(n+\frac 2 {n^{2}}-x)$ for $ n+\frac1 {n^{2}} \leq x \leq n+\frac 2 {n^{2}}$ and $0$ elsewhere. Verify that this is a counter-example.

[$f(n+\frac 1 {n^{2}}) =n \to \infty$]

Answer 2

Counterexample

Set $f_n(x)=n$ for $x\in [0,\frac{1}{2n}]$ and $f(x)=-2n^2x+2n$ for $x\in [\frac{1}{2n}, \frac{1}{n}]$ and $f$ is zero for $x\geq \frac{1}{n}$. Then every $f_n$ is continuous (in $C([0,\infty))$) and the sequence of the integrals goes to $0$ but clearly $f_n$ is not uniformly bounded