If $f$ and $g$ are Riemann integrable, are $f\cdot g$ and $f/g$ Riemann integrable?

If $f,g$ are both Riemann integrable, then $f\cdot g$ is. This follows by proving that if $f$ is Riemann integrable, then $f^2$ is, then using $2fg=(f+g)^2-f^2-g^2$. But we have that $|f^2(x)-f^2(y)|\leqslant 2M|f(x)-f(y)|$ where $M$ is an upper bound for $f$ over the interval in quastion. If $g$ is bounded away from zero, then $f/g$ will be Riemann integrable, for the oscillation of $1/g$ is no greater than that of $g$ times a constant, namely if $g\geqslant m>0$ $$\left|\frac{1}{g(x)}-\frac{1}{g(y)}\right|=\left|\frac{g(x)-g(y)}{g(x)g(y)}\right|\leqslant\frac{1}{m^2}|g(x)-g(y)|$$

It suffices to prove the statement: if $f$ is integrable on $[a,b]$ then so is $f^2$, then $fg = \dfrac 1 2((f + g)^2 - f^2 - g^2) $is integrable. To show that $f^2$ is integrable use the following:

So: $$\begin{align}U(f^2, P) - L(f^2,P) &= \sum (M_i^2 - m_i^2)\Delta x_i \\&< 2T \sum (M_i-m_i)\Delta x_i\\&= 2T(U(f,P) - L(f,P)) \\&<2T\frac{\varepsilon }{2T} = \varepsilon\end{align}$$

Here $T = \sup\limits_{[a,b]} f(x)$. So $f^2$ is integrable.