How can I show the equality of integration for shifting simple functions over $\mathbb{R}$

Question

Let $\phi(x) = \sum_{k=1}^n a_i\chi_{E_i}(x)$ be a simple function on $\mathbb{R}$ with finite support. I want to show that \begin{equation} \int_\mathbb{R} \phi(x) = \int_\mathbb{R} \phi(x+t). \end{equation} It seems rather obvious, since \begin{equation} \sum_{k=1}^n a_i \cdot m(E_i) = \sum_{k=1}^n a_i \cdot m(E_i + t) \end{equation} and furthermore, \begin{equation} m(supp\{x : \phi(x) > 0\}) = m(supp\{x : \phi(x+t) > 0\}) \end{equation} but I am unsure if this is enough to prove my desired statement

Answer 1

In fact I think that it is enough to show that the Lebesgue measure is invariant for "translations of the sets".