Let \(f:[0,1]\to\mathbb R\) be nondecreasing. Assume that \(f\) is differentiable almost everywhere. Prove that \[ \int_0^1 f'(x)\,dx\le f(1)-f(0). \]
Solution
1. Fatou’s lemma bounds the derivative integral by the total increase of \(f\).
For \(0<h<1\), define \[ q_h(x)=\begin{cases} \dfrac{f(x+h)-f(x)}h,&0\le x\le1-h,\\ 0,&1-h<x\le1. \end{cases} \] Since \(f\) is nondecreasing, \(q_h\ge0\). At every point \(x\in[0,1)\) where \(f'(x)\) exists, \[ q_h(x)\longrightarrow f'(x) \qquad(h\downarrow0). \] Choose any sequence \(h_n\downarrow0\). By Fatou’s lemma, \[ \int_0^1 f'(x)\,dx \le \liminf_{n\to\infty}\int_0^1 q_{h_n}(x)\,dx. \] Now \[ \begin{aligned} \int_0^1q_h(x)\,dx &=\frac1h\int_0^{1-h}(f(x+h)-f(x))\,dx\\ &=\frac1h\left(\int_{1-h}^1 f(u)\,du-\int_0^h f(u)\,du\right). \end{aligned} \] Monotonicity gives \[ \int_{1-h}^1 f(u)\,du\le hf(1), \qquad \int_0^h f(u)\,du\ge hf(0). \] Hence \[ \int_0^1q_h(x)\,dx\le f(1)-f(0). \] Combining this with Fatou yields \[ \boxed{\int_0^1 f'(x)\,dx\le f(1)-f(0)}. \]
2. Q.E.D.
Step 1 establishes the required integral inequality.