Product of functions Riemann integrable only if the factors are

Question

For an integral $\displaystyle \int_a^b g(x) f(x) dx$, lets say that $g(x)$ is Riemann integrable and $g(x)f(x)$ is. Can I infer that $f(x)$ is also Riemann integrable? I know that it holds $f,g$ Riemannd integrable $\Rightarrow$ f $\cdot$ g Riemann integrable ... but I'm uncertain of the inversion. Can somebody give a counterexample?

Answer 1

No. Take $[a,b] = [0,1]$ and $g = 0$. For any $f$ you will have $g$ and $fg$ Riemann integrable on $[0,1]$. For example, $f(x)=\frac{1}{x}$ for $x\neq 0$ and $f(0)=0$ will do the trick.

That may seem like cheating. Otherwise you can take $g(x)=x$ and $f(x)=\ln(x)$ for $x\neq 0$ with $f(0)=0$. In this case you are cheating a little less because $g$ is not identically null. If you want to get an example with $g$ never zero you can modify the previous example to get $g$ defined by $g(x)=x$ for $x\neq 0$ with $g(0)=1$ and $f$ given by $f(x)=\ln(x)$ for $x\neq 0$ with $f(0)=0$.

If you want an example with $|g|>\epsilon>0$ on $[a;b]$ you will never find it. In such a case $\frac{1}{g}$ would be bounded on $[a;b]$ and continuous on the same set as $g$. Because $g$ must be Riemann integrable, $\frac{1}{g}$ would be Riemann integrable too. Then $f = \frac{1}{g} \cdot gf$ would be Riemann integrable giving no example.