Integral of the derivative of an increasing function

Question

Let $f : [a, b] \rightarrow \mathbb{R}$ be an increasing function. Then I know that $f'$ is integrable and $$\int_a^b f'(x) dx \leq f(b) - f(a).$$

However, it can be claimed further that

(1) $$\int_a^b f'(x) \leq \lim_{y \rightarrow b^-} f(y) - \lim_{z \rightarrow a+}f(z).$$

(2) Assume further that $f$ is not continuous, then $$\int_a^b f' < f(b) - f(a).$$

I think the result in $(1)$ is stronger than the original version, also for $(2)$ I do not see how to begin the prove, so I try to use contradiction by assuming $\int_a^b f' = f(b) - f(a)$. But I do not see why this force $f$ to be continuous.

The integral here is in Lebesgue sense, and this discussion happen before the Fundamental Theorem of Calculus for Lebesgue integral.

Any suggestion where to begin? Or how the prove should proceed?

Answer 1

For 2), assume that $f$ is not continuous at $x=c$, then \begin{align*} \int_{a}^{b}f'&=\int_{a}^{c}f'+\int_{c}^{b}f'\\ &\leq\lim_{y\rightarrow c^{-}}f(y)-\lim_{z\rightarrow a^{+}}f(z)+\lim_{y\rightarrow b^{-}}f(y)-\lim_{z\rightarrow c^{+}}f(z)\\ &=-(\lim_{z\rightarrow c^{+}}f(z)-\lim_{y\rightarrow c^{-}}f(y))+\lim_{y\rightarrow b^{-}}f(y)-\lim_{z\rightarrow a^{+}}f(z)\\ &<\lim_{y\rightarrow b^{-}}f(y)-\lim_{z\rightarrow a^{+}}f(z)\\ &\leq f(b)-f(a). \end{align*}

Answer 2

As regards (2), if $f$ is not continuous at $c\in(a,b)$ then the left and the right limit exist by monotonicity, they are finite and $$\lim_{x\to b^-}f(x)\leq f(b)\,,\quad\lim_{x\to a^+}f(x)\geq f(a)\,,\quad \lim_{x\to c^-}f(x)<\lim_{x\to c^+}f(x).$$ Hence $$\begin{align}\int_a^b f'(x) dx &=\int_a^c f'(x) dx+\int_c^b f'(x) dx\\ &\leq \left(\lim_{x\to c^-}f(x) - \lim_{x\to a^+}f(x)\right)+\left(\lim_{x\to b^-}f(x)-\lim_{x\to c^+}f(x)\right)\\ &<\lim_{x\to b^-}f(x)-\lim_{x\to a^+}f(x)\leq f(b)- f(a).\end{align}$$ A similar approach holds if $c=a$ or $c=b$.