- \(\star\): Prove that the Lebesgue integral is dilation invariant, i.e. if \(f_\delta(x) = {f({x\over \delta}) \over \delta^n}\) then \(\int f_\delta = \int f\).
Solution
Fix \(\delta > 0\) and let \(m\) be Lebesgue measure on \(\RR^n\).
1. For every measurable \(E \subseteq \RR^n\), \(\int (\chi_E)_\delta = m(E)\).
\((\chi_E)_\delta(x) = \delta^{-n}\chi_E(x/\delta) = \delta^{-n}\chi_{\delta E}(x)\), so \(\int (\chi_E)_\delta = \delta^{-n} m(\delta E) = m(E)\) by the scaling law \(m(\delta E) = \delta^n m(E)\).
2. The claim holds for nonnegative simple \(f = \sum_{i=1}^{m} a_i \chi_{E_i}\).
\(f_\delta = \sum_i a_i (\chi_{E_i})_\delta\), so by linearity of the integral and step 1, \(\int f_\delta = \sum_i a_i m(E_i) = \int f\).
3. The claim holds for nonnegative measurable \(f\).
Choose nonnegative simple functions \(s_k \nearrow f\) pointwise. Then \((s_k)_\delta \nearrow f_\delta\) pointwise, since \((s_k)_\delta(x) = \delta^{-n}s_k(x/\delta)\). By the monotone convergence theorem and step 2, \(\int f_\delta = \lim_k \int (s_k)_\delta = \lim_k \int s_k = \int f\).
4. Q.E.D.
For \(f \in L^1\), \((f_\delta)^\pm = (f^\pm)_\delta\), and step 3 applies to \(f^+\) and \(f^-\); subtracting gives \(\int f_\delta = \int f\).