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Let $f:[0,1] \to \mathbb{R}$ be a monotone increasing function so that $f(0)=0=\lim_{x\to{0^{+}}}f(x)=f(0)$ and $f(1)=1=\lim_{x\to{1^-}}f(x).$ If $\int_0^1f'(t)dt=1$ show that $\int_a^b f'(t)dt=f(b) - f(a)$ for all $0 \leq a < b \leq 1$ and that $f$ is absolutely continuous on $[0,1]$.

Attempt (sketch): I know I can get the second part from the first; roughly, since $f'$ is integrable over $[0,1]$ we have from the absolute continuity of integration that for any finite disjoint union of intervals $I_k=(a_k, b_k)$ in $[0,1]$ with $\sum_k l(i_k)<\delta$, $\int_{\cup I_k}f'=\sum_k \int_{I_k} f'=\sum_k |f(b_k)-f(a_k)| < \epsilon$ for a given $\epsilon>0$, hence proving the absolute continuity of $f$.

My question is how to justify the first part. Again roughly, my routine would be to assume not; therefore, it follows from $f$ monotone increasing that $\int_a^b f'(t)dt < f(b)-f(a)$ for some $(a,b) \in [0,1]$. Thus $\int_0^1 f' = \int_0^a f'+\int_a^b f' + \int_b^1 f'< 1$ contradicting the assumption that $\int_0^1 f'=1$. But in neither part did I use continuity at the left and right endpoints, which makes me feel like I've taken a wrong turn somewhere. Any help is appreciated!

Darrin
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1 Answers1

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Strictly speaking, you haven't justified how you got $\int_0^a f' + \int_a^b f' + \int_b^1 f' < 1$. For this, I think it is enough to just assume $f$ is monotonically increasing, $f(0) = 0$ and $f(1) = 1$.

Claim: Suppose $f$ is monotonically increasing on $[c, d]$. Then $\int_c^d f' \leq f(d) - f(c)$.

Proof: Let $F_n(t) = \frac{f(t + \Delta_n) - f(t)}{\Delta_n}$ where $\Delta_n = (d-c)/n$. Then, $F_n(t)$ converges pointwise almost everywhere to $f'(t)$. So by Fatou's lemma, $$\int_c^d f' \leq \liminf_{n \to \infty} \int_c^d F_n$$

Now note that $$\int_c^d F_n = \sum_{k = 0}^{n-1} \int_{c + k\Delta_n}^{c + (k+1)\Delta_n} F_n \leq \sum_{k=0}^{n-1} [ f(c + (k+2)\Delta_n) - f(c + k \Delta_n) ]$$

As $n \to \infty$, the last sum converges to something $\leq f(d) - f(c)$.