Let $\mathfrak{B}_{\mathbb{R}^n}$ be the Borel $\sigma$-algebra of $\mathbb{R}^n$ and $\lambda :\mathfrak{B}_{\mathbb{R}^n}\to \overline{\mathbb{R}}$ be the Lebesgue measure.
Definition: We say that a function $f:\mathbb{R}^n\to \mathbb{R}$ is $1$-periodic if $f(x+k)=f(x)$ for any $x\in\mathbb{R}^n$ and $k\in\mathbb{Z}^n$.
My question is: Suppose that $f:\mathbb{R}^n\to \mathbb{R}$ is a non-negative measurable function. Does the equality $\int _{[0,1]^n+a}f(x)d\lambda (x)=\int_{[0,1]^n}f(x)d\lambda (x)$ holds for all $a\color{red}{\in\mathbb{R}^n}$ if $f$ is $1$-periodic?
Unfortunately I didn't make any progress.
The only thing is worth mentioning is that we can assume without loss of generality that $a\in [0,1)^n$ because there's $k\in\mathbb{Z}^n$ such that $a-k\in [0,1)^n$ and we have
$\int _{[0,1]^n+a}f(x)d\lambda (x)=\int_{\mathbb{R}^n}\mathbf{1}_{[0,1]^n+a}(x)f(x)d\lambda (x)=$
$\int_{\mathbb{R}^n}\mathbf{1}_{[0,1]^n+a}(x+k)f(x+k)d\lambda (x)=\int_{\mathbb{R}^n}\mathbf{1}_{[0,1]^n+a-k}(x)f(x)d\lambda (x)=$
$\int_{[0,1]^n+a-k}f(x)d\lambda (x)$
Thank you for your attention!