Since $\lim_{n\to\infty} \int_I g(nx)\, dx = 0$ for all intervals $I \subset [0,1]$,we get $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0$$ for any step function $f(x)=\sum_{k=1}^n a_k\chi_{I_k}$ where $I_k\subset[0,1]$ is an interval.

Now suppose $f\in L^1$. Since step functions are dense in $L^1$, for any $\epsilon>0$ we can write $f=f_1+f_2$ where $f_1$ is a step function and $\int_0^1|f_2|\,dx<\epsilon$. Using what we've proven for step functions and the boundedness of $g$ gives that $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0.$$