Suppose that \(f\in L^1([0,1])\) and define \[ g(x)=\int_x^1\frac{f(t)}{t}\,dt. \] Show that \(g\in L^1([0,1])\) and that \[ \int_0^1 g(x)\,dx=\int_0^1 f(x)\,dx. \]
Solution
Consider first the nonnegative integrand \[ H(x,t)=\mathbf 1_{\{0\le x\le t\le1\}}\frac{|f(t)|}{t}. \] Tonelli’s theorem gives \[ \begin{aligned} \int_0^1\int_x^1\frac{|f(t)|}{t}\,dt\,dx &=\int_0^1\int_0^t\frac{|f(t)|}{t}\,dx\,dt\\ &=\int_0^1 |f(t)|\,dt<\infty. \end{aligned} \] Hence for almost every \(x\) the defining integral for \(g(x)\) is absolutely convergent, and moreover \[ \int_0^1|g(x)|\,dx \le \int_0^1\int_x^1\frac{|f(t)|}{t}\,dt\,dx \le \|f\|_1. \] Thus \(g\in L^1([0,1])\).
Since the corresponding signed integrand is absolutely integrable on the triangle \(0\le x\le t\le1\), Fubini’s theorem applies and yields \[ \begin{aligned} \int_0^1g(x)\,dx &=\int_0^1\int_x^1\frac{f(t)}{t}\,dt\,dx\\ &=\int_0^1\int_0^t\frac{f(t)}{t}\,dx\,dt\\ &=\int_0^1 f(t)\,dt. \end{aligned} \] This proves both claims.