Let \(f\in L^1(\mathbb R)\) and define \[ h(x)=\int_x^{x+1}f(t)\,dt. \]
Prove that \(h\) is absolutely continuous.
Prove that \(h(x)\to0\) as \(x\to\infty\).
Solution
Define \[ F(x)=\int_0^x f(t)\,dt, \] with the usual signed interpretation when \(x<0\). Since \(f\in L^1(\mathbb R)\), the indefinite integral \(F\) is absolutely continuous, and \[ F'(x)=f(x) \] for almost every \(x\). Since \[ h(x)=F(x+1)-F(x), \] the function \(h\) is a difference of absolutely continuous functions and is therefore absolutely continuous. In fact, \[ h'(x)=f(x+1)-f(x) \] for almost every \(x\).
For part (b), \[ |h(x)|\le\int_x^{x+1}|f(t)|\,dt\le\int_x^\infty|f(t)|\,dt. \] Because \(f\in L^1(\mathbb R)\), the tail integral tends to zero as \(x\to\infty\). Hence \[ h(x)\longrightarrow0. \]