$f \in L^1 $, does that imply $f(\lfloor x \rfloor) \in L^1$? As the title says, I am working on a problem and I would like to know if there is a relation between a function $f$ being integrable and the function $f(\lfloor x \rfloor)$ being integrable. One can show that 
$$\int|f(x) - f(\lfloor x \rfloor)|dx  + \int |f(x)|dx \geq \int|f(\lfloor x \rfloor)|dx $$
so if the first integral on the left converges, then $f(\lfloor x \rfloor)$ must be integrable. It is true that the integrand is bounded a.e, but more than that I have not been able to show. 
Is the statement even true, and if so any hints on how to show this? 

Consider the function $\Phi(x)=\begin{cases}0&\text{if }\lvert x\rvert\ge1\\ \exp\frac1{x^2-1}&\text{if }\lvert x\rvert<1\end{cases}$ and $\phi(x)=\frac{\Phi(t)}{\int_{-1}^1\Phi(t)\,dt}$. Notice that $\int_\Bbb R\phi(t)\,dt=1$ and $\max_{t\in \Bbb R} \phi(t)=\phi(0)>0$.
Consider $\phi_n(t)=\phi(n^2(t-n))$. This function satisfies $\int_{\Bbb R}\phi_n(t)\,dt=\frac1{n^2}\int_{\Bbb R}\phi(t)\,dt=n^{-2}$. Its maximum is $\phi_n(n)=\phi(0)$ and its support is the interval $[n-n^{-2},n+n^{-2}]$.
The function $f(x)=\sum\limits_{n= 2}^\infty\phi_n(x)$ satisfies $$\int_{\Bbb R}\lvert f(t)\rvert\,dt=\int_{\Bbb R}f(t)\,dt=\sum_{k\ge2}\frac1{k^2}<\infty\\ f(\lfloor x\rfloor)=\begin{cases}\phi(0)&\text{if }x\ge 2\\ 0&\text{if }x<2\end{cases}$$
This makes it a counterexample to your claim.
