Prove that $f(x) - f(a) = \int_{a}^{x} f^{'}, \forall x \in [a,b].$

Question

Assume that $f:[a,b]\to\mathbb R$ has a finite derivative for every $x$, and that $f^{'}$ is Lebesgue integrable over $[a,b].$ Prove that $$f(x) - f(a) = \int_{a}^{x} f'(t)\,dt, \,\,\,\text{for all}\,\, x \in [a,b].$$

Hint given to me:

Define the function $g_{n}(x)$ (called cutting from above function)as follows:

$$g_{n}(x) = \begin{cases} f'(x), & \textbf{if} \quad f'(x) \leq n, \\ n, & \textbf{if} \quad f'(x) > n. \end{cases}$$

And define $h_{n}(x)$ (called cutting from below function)as follows:

$$h_{n}(x) = \begin{cases} f'(x), & \textbf{if} \quad f'(x) \leq n, \\ -n, & \textbf{if} \quad f'(x) > n. \end{cases}$$

But still, I did not get what next should I do to write a rigorous solution could anyone help me in this, please?

Answer 1

Set $g_n(x)=n\big( f(x+\tfrac{1}{n})-f(x)\big)$. Then, for every $t\in [a,x]$ and $n\in\mathbb N$, there exists an $\xi_n$, such that $$ |g_n(x)|=|f'(\xi_n)|\le \sup |f'(x)|=M<\infty. $$

Then $g_n(x)\to f'(x)$ pointwise and $|g_n(x)|\le M$. Hence by virtue of the Lebesgue Dominated Convergence Theorem $$ \lim_{n\to\infty}\int_a^x g_n(x)\,dx=\int_a^x f'(x)\,dx. $$ But $$ \int_a^x g_n(x)\,dx=n\int_b^{b+1/n} f(t)\,dt-n\int_a^{a+1/n} f(t)\,dt\to f(x)-f(a). $$

Answer 2

Theorem 1, p. 279, M. Spivak, Calculus (3rd Edition):

If $g: [a,b]\to\mathbb R$ is Riemann integrable, then $$ \lim_{\|P\|\to 0}S(f,P,\boldsymbol{\xi})=\int_a^b f(x)\,dx $$ where $P$ runs over all the partitions of $[a,b]$ and $\boldsymbol{\xi}=\{a<\xi_1<\cdots <x_{n-1}<b\}$ is any choice of points in the subintervals of the partition. If $P=\{a<t_1<\cdots<t_{n-1}<b\}$, then $\xi_j\in[x_{j-1},x_j]$.

Now, if $P=\{a=t_0<t_1<\cdots<t_{n-1}<t_n=x\}$ is a partition of $[a,x]$

$$ f(x)-f(a)=\sum_{i=1}^n \big(f(t_i)-f(t_{i-1})\big) =\sum_{i=1}^n f'(\xi_i)(t_i-t_{i-1})=S(f',P,\boldsymbol{\xi})\to\int_a^xf'(t)\,dt $$