Integral convergence implies pointwise This is a simplified version of a question I am looking at, but embarrassingly I can't do this one anyway. 
Let's assume $f(x)$ is a decreasing positive function, $f(0)$ is infinite and, moreover, $f(x)$ is comparable to $1/x$ near $x=0$. We look only at positive $x$'s.
Question. Assume we know that the limit
$$
\lim_{a\to 0^+}\int\limits_{a}^{2a} f(t)dt
$$
exists as a positive finite number. Does it imply that
$$
\lim_{n\to\infty} f(2^{-n})\cdot 2^{-n}
$$
exists? 
Note: it definitely does not imply that the limit of $xf(x)$ exists. 
 A: If you are ok with $f$ being not continuous and not strictly decreasing, here's a simple enough counter-example. (With a bit more work you can make it continuous and strict, but it distracts from the point.)
Let $f(x)$ be the step-function defined by 
$$ f(x) = \begin{cases}
2^{k} & x \in ( 2^{-k}, 2^{1-k}) \\
2^{k} & x = 2^{-k}, \quad k \text{ is odd} \\
2^{k+1} & x = 2^{-k}, \quad k\text{ is even}
\end{cases} $$
You have that the integral 
$$ \int_a^{2a} f(t) ~dt  = 1 $$
for any $a$. 
You have that $2^{-k} f(2^{-k})$ alternating between 1 and 2 and does not converge. 

Incidentally, the "comparability to $1/x$" in your question is superfluous. Since you assumed decreasing, you have that 
$$ a f(2a) \leq \int_a^{2a} f(x) ~dx \leq a f(a) $$
which implies that 
$$ \liminf xf(x) \geq \lim \int_a^{2a} f(x) ~dx $$
$$ \limsup xf(x) \leq 2 \lim\int_a^{2a} f(x)~ dx$$ 
so comparability is automatic, and also the existence of a sequence $x_n \to 0$ such that $x_n f(x_n)$ converges. (Of course $(2^{-n})$ need not be such a sequence.)
