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.