Continuity of $f(x) (a^{[x]})$ Lets assume $a$ is a constant and $f(x) = cos(\pi x)$, and $[x]$ is the greatest integer function of $x$. As far as I know $[x]$ is a discontinuous function. 
Now my question:
$g(x) = f(x) (a^{[x]})$
Where $g:\mathbb{R}\to \mathbb{R}$. Is $g(x)$ a continuous function? Meaning is it possible to do an indefinite integral to $g(x)$? If it is, how would that be done?
 A: No, $g(x)$ is not necessarily continuous unless $f(x)=0$ for $x\in\mathbb{Z}$. However one can find the indefinite integral provided $f$ is integrable.
Suppose $a>0$ and $f$ is defined and integrable for $x\ge0$
Then
\begin{eqnarray}
\int_0^t a^{\lfloor x\rfloor}f(x)\,dx&=&\int_0^1f(x)\,dx+a\int_1^2f(x)\,dx+\cdots+a^{\lfloor t\rfloor-1}\int_{\lfloor t\rfloor-1}^{\lfloor t\rfloor}f(x)\,dx+a^{\lfloor t\rfloor}\int_{\lfloor t\rfloor}^tf(x)\,dx\\
&=&\sum_{k=0}^{\lfloor t\rfloor-1}\left[a^k\int_k^{k+1}f(x)\,dx\right]+a^{\lfloor t\rfloor}\int_{\lfloor t\rfloor}^tf(x)\,dx
\end{eqnarray}
To express the antiderivative as a function of $x$, let $t$ be the integration variable, then
$$ F(x)=\int_0^x a^{\lfloor t\rfloor}f(t)\,dt= \sum_{k=0}^{\lfloor x\rfloor-1}\left[a^k\int_k^{k+1}f(t)\,dt\right]+a^{\lfloor x\rfloor}\int_{\lfloor x\rfloor}^xf(t)\,dt$$
and 
$$ \int_{x_0}^{x_1}a^{\lfloor t\rfloor}f(t)\,dt=F(x_1)-F(x_0) $$
Note that this result is not as nice and compact as the antiderivative of the floor function because, unlike the floor function, $f$ is not constant on intervals $[k,k+1)$ for $k\in\mathbb{Z}$.
