Suppose $f (x)$ is bounded and has a primitive on $[a,\,b]$, and $g\in C \big([a,\,b]\big) $. Must $f (x)\cdot g (x)$ have a primitive? Suppose $f (x)$ is bounded and has a primitive on $[a,\,b]$, and $g\in C \big([a,\,b]\big) $. Must $f (x)\cdot g (x)$ have a primitive?


Suppose it has a primitive, then it's necessary to be Riemann integrable in $[a,\,b]$.
With these hypotheses it is true that $fg$ has a primitive.

For $f$ to have a primitive on $[a,b]$ means that there exists a differentiable function $F$ such that $F'(x) = f(x)$ for all $x \in [a,b]$.

We first show that the function $x \mapsto xf(x)$ must also have a primitive. Taking $G(x) = xF(x)$ we have $G'(x) = xF'(x) + F(x) = xf(x) +F(x)$. Since $F$ is differentiable it is continuous and Riemann integrable on $[a,b]$, and, therefore, has a primitive $H' = F$. Thus $xf(x) = G'(x) - H'(x)$ and $x \mapsto xf(x)$ has a primitive.  It follows by induction that $x \mapsto p(x)f(x)$, where $p(x)$ is a polynomial, must have a primitive.

Since $g$ is continuous, it follows by the Weierstrass approximation theorem that there exists a sequence of polynomials $(p_n)$ that converges uniformly to $g$. Since $f$ is bounded, it follows that $p_nf$ converges uniformly to $fg$.

Finally, we apply the following lemma:


  If $h_n \to h$ uniformly and each $h_n$ has a primitive, then $h$ has
  a primitive.


Since each function $p_nf$ has a primitive and in view of the uniform convergence shown above, it follows that $fg$ has a primitive.
Alternatively consider 

$$
H(x) = \int_a^x f(t)g(t)\, dt 
$$

Then since $f$ has a primitive it is integrable, $g$ is bounded and continuous so the product $fg$ is integrable and $H$ is well-defined.