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 \(m(E + h) = m(E)\).
1.1. \(m^*(A + h) = m^*(A)\) for every \(A \subseteq \RR^n\).
Translation by \(h\) is a bijection between countable covers of \(A\) by closed boxes and countable covers of \(A + h\) by closed boxes, and it preserves the volume of each box. Taking infima gives equality.
1.2. \(E + h\) is measurable.
1.1. So \(E + h = (G + h) \setminus (Z + h)\) is measurable.
1.3. Q.E.D.
Step 2 gives measurability, and on measurable sets \(m = m^*\), so step 1 gives \(m(E + h) = m(E)\).
2. For measurable \(f \geq 0\), \(\tau_h f\) is measurable and \(\int \tau_h f = \int f\).
2.1. \(\tau_h f\) is measurable.
For every \(a \in \RR\), \(\theset{\tau_h f > a} = \theset{f > a} + h\), which is measurable by step 1.
2.2. For measurable \(E\), \(\int \tau_h \chi_E = m(E)\).
\(\tau_h\chi_E(x) = \chi_E(x - h) = \chi_{E + h}(x)\), and \(m(E + h) = m(E)\) by step 1.
2.3. Q.E.D.
By linearity, step 2 gives \(\int \tau_h s = \int s\) for nonnegative simple \(s\). Choose simple \(0 \leq s_k \nearrow f\). Then \(\tau_h s_k \nearrow \tau_h f\), and the monotone convergence theorem gives \(\int \tau_h f = \lim_k \int \tau_h s_k = \lim_k \int s_k = \int f\).
3. Q.E.D.
Steps 1 and 2 are the two parts.