Let $f$ be monotonic increasing on $[a,b]$ If $\int_a^bf'(x)dx = f(b)-f(a)$ Show $f$ is absolutely continuous

Question

I tried to prove it by the definition of absolutely continouse:

Let $\{(a_i,b_i)\}_{i=0}^n$ be disjoint sub-interval on $[a,b]$

By Reimenn integral,

$\int_a^bf'(x)dx = \sum_{i=0}^nf'(t_i)(b_i-a_i)$ ,for any $t_i \in[a_i,b_i]$

Since f is differentiable, there exist $c_i\in[a_i,b_i] $ such that $\frac{f(b_i)-f(a_i)}{b_i-a_i} = f'(c_i)$ ......

And then I think I am on the wrong direction. What else can I know from $\int_a^bf'(x)dx = f(b)-f(a)$

Help me please

Answer 1

Extend $f$ to be $f(a)$ for $x\leq a$ and $f(b)$ for $x\geq b$. Since $f$ is increasing, for every $h>0$ we have% \begin{align*} \int_{a}^{b}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\,dx & =n{\biggl \{}% \int_{b}^{b+\frac{1}{n}}f(x)\,dx-\int_{a}^{a+\frac{1}{n}}f(x)\,dx{\biggr \}}\\ & \leq n\frac{f(b)-f(a)}{n}=f(b)-f(a). \end{align*} It follows from Fatou's lemma, that $$ 0\leq\int_{a}^{b}f^{\prime}(x)\,dx\leq\liminf_{n\rightarrow\infty}\int_{a}% ^{b}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\,dx\leq f(b)-f(a). $$ With the same proof if $a<y<b$ you get \begin{align*} \int_{a}^{y}f^{\prime}(x)\,dx & \leq f(y)-f(a),\\ \int_{y}^{b}f^{\prime}(x)\,dx & \leq f(b)-f(y). \end{align*} Adding these two inequalities gives $$ \int_{a}^{y}f^{\prime}(x)\,dx+\int_{y}^{b}f^{\prime}(x)\,dx\leq f(y)-f(a)+f(b)-f(y)=f(b)-f(a). $$ Since by hypothesis $\int_{a}^{b}f^{\prime}(x)\,dx=f(b)-f(a)$, it follows that both inequalities must be equalities, that is, $$ f(y)=f(a)+\int_{a}^{y}f^{\prime}(x)\,dx $$ for every $y\in\lbrack a,b]$. Now you need to prove that if $g\geq0$ is integrable, then the function $$ h(y)=\int_{a}^{y}g(x)\,dx $$ is absolutely continuous.

Answer 2

the function is differentiable on that interval, so it has to be continuous. and continuous functions are absolutely continuous on an interval. you could use the definition of absolute continuity to prove that a continuous function is absolutely continuous on an interval.