I guess not. Because we usually require $|f|$ to be integrable on ℝ so that it has the fourier transform.
Can anyone give me an counterexample for the statement in the title? I have searched for quite a while but still haven't got an counterexample.
I am actually trying to find the necessary and sufficient condition so that the following equality holds. $$F(b)−F(a)=\int_{\mathbb{R}}\frac{e^{−iat}−e^{−ibt}}{it}\phi (t)dt,\forall a,b \in \mathbb{R} $$ where $F$ is the cumulative distribution function of a random variable $X$ and $\phi$ is the characteristic function of $X$. The integral is in the Lebesgue sense.
I found that $\frac{e^{−iat}−e^{−ibt}}{it}\phi(t)$ integrable for all $a,b$ and $X$ does not have point mass would be a necessary and sufficient condition. But it is hard to check $\frac{e^{−iat}−e^{−ibt}}{it}\phi(t)$ integrable for all $a,b$.
Since it is not hard to see that $\int_{[T,∞)∪(−∞,−T]}|\phi(t)|/|t|dt<∞ $for some $T>0$ would imply that $\frac{e^{−iat}−e^{−ibt}}{it}\phi(t)$ integrable for all $a,b$. By taking $a=−1,b=1, \frac{e^{−iat}−e^{−ibt}}{it}\phi(t)$ integrable for all $a,b$ implies $2\sin(t)\phi(t)/t$ integrable and then I am wondering whether the statement in the title is true.
As commenters said, the answer is negative: e.g., $f(x) =|x|^{-1} e^{-|x|}$ is a nonintegrable function such that $|f(x)\sin x|$ is globally integrable.
More generally, the following are equivalent:
Indeed, if 2) fails then the sets $A_n = \{x: 2^{-n}\le |g(x)|<2^{1-n}\}$ have positive measure for infinitely many $A_n$. Restrict attention to only such $n$ and pick a subset $B_n\subset A_n$ of finite positive measure for each of them. Then the function
$$ f(x) = \sum_n |B_n|^{-1} \chi_{B_n} $$ is not integrable, but $fg$ is. $\quad\Box$
The implication $2) \implies 1)$ is easy.