Remains true for f integrable? $f(x)=f(x+1)\ \forall x\Rightarrow \int_0^1f(x+t)dt=\int_0^1f(t)dt$, when $f$ is continuous on $[0,1]$. The proof it is not hard. My question is, this property remains true if $f$ is only integrable? 

Given $x\in\Bbb R$, write $k=\lfloor x\rfloor$ and $y=x-k$, so that $0\le y<1$. Then $f(u+k)=f(u)$ for all $u$, so
$$
\int_0^1 f(t+x)\,dt = \int_0^1 f(t+y+k)\,dt = \int_0^1 f(t+y)\,dt.
$$
Furthermore,
\begin{align*}
\int_0^1 f(t+y)\,dt &= \int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y)\,dt\\ &= \int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y-1)dt
\end{align*}
using $f(u)=f(u-1)$. Now make the change of variables $u=t+y$ in the first integral and $u=t+y-1$ in the second integral:
$$
\int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y-1)\,dt = \int_y^1 f(u)\,du + \int_0^y f(u)\,du = \int_0^1 f(u)\,du.
$$
