Is a function bounded if it has an antiderivative and converges as its argument grows without bound? Main question:
Let $f : [0,\infty) \to \mathbb R$ have an antiderivative. Furthermore, suppose $\underset {t \to \infty} \lim f(t) = 0$. Can I conclude that $f$ is bounded?
If $f$ were continuous, the answer would be “yes”, for obvious reasons. But, with just the given information, I find it difficult to tell.

Motivation:
This is given solely for context. Please do not answer the question here! I am trying to solve the following problem. Let $A$ be a square matrix whose eigenvalues all have negative real part. Let $f : \mathbb R \to \mathbb C^n$ have an antiderivative, and suppose $\underset {t \to \infty} \lim f(t) = 0$. Show that
$$\underset {t \to \infty} \lim \int_0^t e^{(t-s)A} g(s) ds = 0$$
 A: Take
$$f(x) = \begin{cases}0, \quad x = 0\\ 2x \sin \frac{1}{x^2} - \frac{2}{x} \cos \frac{1}{x^2}, \quad 0 < x \leqslant1 \end{cases}$$
and extend smoothly to $[0,\infty)$ such that $f(x) \to 0 $ as $x \to \infty$.
The function has an antiderivative with $F'(x) = f(x)$ such that on $[0,1]$ we have
$$F(x) = \begin{cases}0, \quad x = 0 \\ x^2 \sin\frac{1}{x^2}, \quad 0 < x \leqslant 1 \end{cases}$$
but $f$ is unbounded near $x = 0$.
A: NO. We can find a differentiable $F:\Bbb R\to \Bbb R$ with $F(x)=0$ for $x\leq 0 $ and $F(x)=1$ for $x\geq 0,$ such that $\{F'(x): 0<x<1\}$ is unbounded. Then $f=F'$ has an anti-derivative, and $f'(x)=0$ for $x\not \in (0,1),$ but $f$ is unbounded on $(0,1).$
(i). (Exercise).For  $a<b$ and $c<d$ and for  any $r>0$ there exists a monotonic differentiable $F:[a,b]\to [c,d]$  with $F(a)=c$ and $F(b)=d$  and $F'(a)=F'(b)=0,$ such that $\sup \{F'(x): x\in (a,b)\}>r.$ 
(ii). For $n\in \Bbb N$ let $a_n=1-2^{-n}$ and $b_n=1-3^{-n}.$ By (i)  there exists  a monotonic differentiable $F:[a_n,a_{n+1}]\to [b_n,b_{n+1}]$ with $F(a_n)=b_n$ and $F(a_{n+1})=b_{n+1} $  and $F'(a_n)=F'(a_{n+1)})=0$, such that $\sup \{F'(x):x\in (a_n,a_{n+1})\}>n.$  
And let $F(x)=0$ for $x<0$ and $F(y)=1$ for $y\geq 1.$
It remains to show that $F'(1)$ exists and is equal to $0.$ It suffices to show that $\lim_{x\to 1^-}\frac {1-F(x)}{1-x}=0.$ For $0<x<1$ there exists a unique $n_x\in \Bbb N$ such that   $ a_{n_x}\leq x<a_{(n_x+1)}.$ 
So $0<\frac {1-F(x)}{1-x}<$ $ \frac {1-F(a_{n_x})}{1-a_{(n_x+1)}}=$ $\frac {3^{-n_x}}{2^{-n_x-1}},$ which $\to 0$ as $n_x\to \infty.$ And $n_x\to \infty$ as $x\to 1^-.$
A: No, one cannot so conclude. Let me add (a somewhat artificial) counterexample to the existing list:
Consider the hyperbolic branch $1/x$ in the first quadrant of $\mathbb R^2$, and define the function $f$ with $f(x)=\alpha$ for some $\alpha\in\mathbb R$ when $x=0$ and $f(x)=1/x$ for $x\in(0,\infty)$.
PS. You can only conclude that a function that satisfies your conditions is either bounded above or bounded below.
