Prove that if \(E \subseteq \RR^n\) is a Lebesgue measurable set, then for any \(h \in \RR\) the set \[ E+h \da \ts{x + h \st x\in E } \] is also Lebesgue measurable and satisfies \(m(E + h) = m(E)\).
Prove that if \(f\) is a non-negative measurable function on \(\RR^n\) and \(h\in \RR^n\) then the function \[ \tau_h d(x) \da f(x-h) \] is a non-negative measurable function and \[ \int f(x) \dx = \int f(x-h) \dx .\]
Solution
Take \(h \in \RR^n\) and write \(\tau_h f(x) \coloneqq f(x-h)\).
1. For measurable \(E\), \(E + h\) is measurable and \(m(E + h) = m(E)\).
Translation by \(h\) is a bijection between countable covers of a set \(A\) by closed boxes and countable covers of \(A + h\), preserving each volume, so \(m^*(A + h) = m^*(A)\) for every \(A\). Write \(E = G \setminus Z\) with \(G\) a \(G_\delta\) set and \(Z\) null. Translation is a homeomorphism, so \(G + h\) is \(G_\delta\), and \(m^*(Z + h) = 0\). So \(E + h = (G + h) \setminus (Z + h)\) is measurable, and \(m(E + h) = m^*(E + h) = m^*(E) = m(E)\).
2. For measurable \(f \ge 0\), \(\tau_h f\) is measurable and \(\int \tau_h f = \int f\).
1. For \(f = \chi_E\), \(\tau_h\chi_E = \chi_{E+h}\) has integral \(m(E + h) = m(E)\) by step 1, and linearity extends this to nonnegative simple functions. For general \(f\), choose simple \(0 \le s_k \uparrow f\); then \(\tau_h s_k \uparrow \tau_h f\), and the monotone convergence theorem gives \(\int \tau_h f = \lim_k \int s_k = \int f\). See P-5CM5W.