- \(\star\): Prove continuity in \(L^1\), i.e. \[ f \in L^{1} \Longrightarrow \lim _{h \rightarrow 0} \int|f(x+h)-f(x)|=0 .\]
Solution
1. For a bounded interval \(I\) and \(h \in \RR\), \(\int |\chi_{I}(x+h) - \chi_{I}(x)|\,dx \leq 2|h|\).
The integral is the measure of the symmetric difference of \(I\) and \(I - h\), and each endpoint of \(I\) contributes a set of measure at most \(|h|\) to it.
2. For every step function \(s = \sum_{i=1}^k c_i \chi_{I_i}\) with bounded intervals \(I_i\), \(\lim_{h\to0}\int |s(x+h) - s(x)|\,dx = 0\).
By the triangle inequality and step 1, \(\int |s(x+h) - s(x)|\,dx \leq \sum_i |c_i| \int |\chi_{I_i}(x+h) - \chi_{I_i}(x)|\,dx \leq 2|h|\sum_i |c_i|\).
3. Step functions are dense in \(L^1(\RR)\).
Simple functions \(\sum_i c_i\chi_{E_i}\) with \(m(E_i)<\infty\) are dense in \(L^1(\RR)\), so it suffices to approximate \(\chi_E\) for \(m(E) < \infty\). By outer regularity there is an open \(U \supseteq E\) with \(m(U \setminus E) < \eps/2\). \(U\) is a countable disjoint union of open intervals \(J_k\) with \(\sum_k m(J_k) = m(U) < \infty\), so for some \(N\), \(A = \bigcup_{k\le N} J_k\) satisfies \(m(U \setminus A) < \eps/2\). Then \(\norm{\chi_E - \chi_A}_1 = m(E \triangle A) < \eps\).
4. For every \(f \in L^1(\RR)\) and \(\eps > 0\) there is \(\delta > 0\) with \(\int |f(x+h) - f(x)|\,dx < \eps\) for \(|h| < \delta\).
By step 3 choose a step function \(s\) with \(\norm{f - s}_1 < \eps/3\), and by step 2 choose \(\delta > 0\) with \(\int |s(x+h) - s(x)|\,dx < \eps/3\) for \(|h| < \delta\). By the triangle inequality and translation invariance of Lebesgue measure, \[ \int |f(x+h) - f(x)|\,dx \leq \int |f(x+h) - s(x+h)|\,dx + \int |s(x+h) - s(x)|\,dx + \int |s(x) - f(x)|\,dx < \eps. \]
5. Q.E.D.
Step 4 is the definition of \(\lim_{h \to 0} \int|f(x+h)-f(x)|\,dx = 0\).