- \(\star\): Prove that the Lebesgue integral is translation invariant, i.e. if \(\tau_h(x) = x+h\) then \(\int \tau_h f = \int f\).
Solution
Write \(\tau_h f \coloneqq f \circ \tau_h\), so \(\tau_h f(x) = f(x+h)\), and let \(m\) be Lebesgue measure on \(\RR^n\).
1. 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 translation invariance of Lebesgue measure.
2. The claim holds for nonnegative simple \(s = \sum_i a_i\chi_{E_i}\).
By linearity and step 1, \(\int \tau_h s = \sum_i a_i m(E_i) = \int s\).
3. The claim holds for measurable \(f \geq 0\).
Choose simple functions \(0 \leq s_k \nearrow f\) pointwise. Then \(\tau_h s_k \nearrow \tau_h f\) pointwise, and the monotone convergence theorem with step 2 gives \(\int \tau_h f = \lim_k \int \tau_h s_k = \lim_k \int s_k = \int f\).
4. Q.E.D.
For \(f \in L^1\), \((\tau_h f)^\pm = \tau_h(f^\pm)\); apply step 3 to \(f^+\) and \(f^-\) and subtract.