Let $f,w$ be positive decreasing functions on $(0,\infty)$ with $\int_0^\infty w(t)dt =\infty$ and $\int_0^\infty f(t)w(t) dt <\infty$. Can we say that $\int_0^\infty \log(f(t)+1) <\infty$? or do we have a counterexample?
It is clear that $\int_0^\infty \log(f(t)w(t)+1) <\infty$ since $ \log(f(t)w(t)+1) \le C f(t)w(t) $ for all $t$ for some $C$ (note that $\log (t+1) \le C t$). Of course, when $\int_0^\infty w(t)dt <\infty$, it is very easy to find a counterexample.
No. Consider $f(x)=w(x)=\min\{1,\frac1x\}$. Then, $\log(f(x)+1)\approx \frac1x$ as $x\to\infty$ and thus the integral diverges. On the other hand $\int_0^\infty f(x)w(x)\,dx=2$